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mJackiemJackie
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20190106
1 parent 1f63d60 commit 8c3e41f

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‎code/lc160.java

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package code;
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/*
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* 160. Intersection of Two Linked Lists
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* 题意：求两个链表的交叉点
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* 难度：Easy
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* 分类：LinkedList
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* 思路：两种方法：1.找出两个链表的长度差x，长的先走x步； 2.走完一个链表，走另一个链表，两个cur都走了两个链表长度的和步
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* Tips：两种方法的本质是一样的其实，都是找到了步数差，都遍历了两遍
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*/
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public class lc160 {
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public class ListNode {
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int val;
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ListNode next;
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ListNode(int x) { val = x; }
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}
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public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
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if(headA==null||headB==null)
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return null;
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ListNode curA = headA;
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ListNode curB = headB;
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while (curA != curB) {
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if(curA == null)
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curA = headB;
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if(curB == null)
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curB = headA;
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if(curA==curB)
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return curA; //别忘了判断一下是否相等
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curA = curA.next;
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curB = curB.next;
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}
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return curA;
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}
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}

‎code/lc169.java

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package code;
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/*
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* 169. Majority Element
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* 题意：数组中有一个元素出现次数 >len/2 ,找出这个数
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* 难度：Easy
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* 分类：Array, Divide and Conquer, Bit Maniputation
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* 思路：很多种方法， Hashmap 是 O(n), O(n)的。 快排是O(log(n)), O(1)的。最巧妙的办法是 O(n), O(1) 的如下。
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* Tips：之所以能够 O(n), O(1) 是因为题目已经给定了数组中一定能找到这个数，该方法充分利用了这一点
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*/
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public class lc169 {
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public int majorityElement(int[] nums) {
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int res = nums[0];
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int count = 1;
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for (int i = 1; i < nums.length ; i++) {
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if(nums[i]!=res) // 不是这个数就 --， ==0就用当前数替换res
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count--;
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else
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count++;
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if(count==0){
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res = nums[i];
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count++;
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}
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}
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return res;
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}
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}

‎code/lc198.java

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package code;
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/*
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* 198. House Robber
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* 题意：数组最大和，不能选取相邻的两个数
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* 难度：Easy
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* 分类：Dynamic Programming
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* 思路：经典的dp题，记一下
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* Tips：时间复杂度为 O(n)
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*/
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public class lc198 {
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public int rob(int[] nums) {
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if(nums.length == 0)
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return 0;
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if(nums.length == 1)
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return nums[0];
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int[] dp = new int[nums.length];
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dp[0] = nums[0];
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dp[1] = Math.max(nums[0], nums[1]);
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for (int i = 2; i < nums.length ; i++) {
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dp[i] = Math.max((dp[i-2] + nums[i]),dp[i-1]); //dp[i] 表示以 0~i 的数组的结果
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}
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return dp[nums.length-1];
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}
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}

‎code/lc200.java

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package code;
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/*
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* 200. Number of Islands
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* 题意：0,1矩阵，求1片区个数
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* 难度：Medium
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* 分类：Depth-first Search, Breadth-first Search, Union Find
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* 思路：两种方法，一种搜索，一种并查集
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* Tips：
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*/
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public class lc200 {
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public int numIslands(char[][] grid) {
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if(grid.length==0)
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return 0;
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int count = 0;
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for (int i = 0; i < grid.length ; i++) {
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for (int j = 0; j < grid[i].length ; j++) {
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if( grid[i][j]=='1' ){
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count++;
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grid[i][j] = '0'; //置0， 下次不用考虑
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search(grid,i,j);
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}
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}
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}
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return count;
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}
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public void search(char[][] grid, int i, int j){
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if( i>0 && grid[i-1][j]=='1') {
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grid[i-1][j] = '0'; //置0， 下次不用考虑
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search(grid, i - 1, j);
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}
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if( j>0 && grid[i][j-1]=='1') {
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grid[i][j-1] = '0';
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search(grid, i, j-1);
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}
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if( i+1<grid.length && grid[i+1][j]=='1') {
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grid[i+1][j] = '0';
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search(grid, i + 1, j);
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}
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if( j+1<grid[0].length && grid[i][j+1]=='1') {
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grid[i][j+1] = '0';
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search(grid, i, j+1);
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}
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return;
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}
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//并查集
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int[] flag;
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int count = 0;
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public int numIslands2(char[][] grid) {
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if(grid.length==0)
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return 0;
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//初始化， 每个1都是一个island
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flag = new int[grid.length*grid[0].length];
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for (int i = 0; i < grid.length ; i++) {
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for (int j = 0; j < grid[0].length ; j++) {
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if(grid[i][j]=='1'){
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int id = i*grid[0].length+j;
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flag[id] = id;
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count++;
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}
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}
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}
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for (int i = 0; i < grid.length ; i++) {
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for (int j = 0; j < grid[0].length ; j++) {
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if(grid[i][j]=='1'){
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int id1 = i*grid[0].length+j;
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if( i>0 && grid[i-1][j]=='1') { //四个方向都合一下，虽然有些冗余
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int id2 = (i-1)*grid[0].length+j;
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union(id1, id2);
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}
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if( j>0 && grid[i][j-1]=='1') {
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int id2 = i*grid[0].length+j-1;
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union(id1, id2);
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}
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if( i+1<grid.length && grid[i+1][j]=='1') {
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int id2 = (i+1)*grid[0].length+j;
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union(id1, id2);
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}
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if( j+1<grid[0].length && grid[i][j+1]=='1') {
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int id2 = i*grid[0].length+j+1;
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union(id1, id2);
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}
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}
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}
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}
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return count;
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}
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public void union(int id1, int id2){
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int flag_id1 = find(id1);
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int flag_id2 = find(id2);
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if(flag_id1==flag_id2)
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return;
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else{
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flag[flag_id1] = flag[flag_id2]; //把father们的 flag_id 合一起了
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count--;
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}
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}
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public int find(int id){ //找到father的flag_id
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int res = flag[id];
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while(res != id){ //接着往上找father
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id = res;
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res = flag[id];
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}
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return res;
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}
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}

‎code/lc206.java

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package code;
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/*
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* 206. Reverse Linked List
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* 题意：链表反转
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* 难度：Easy
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* 分类：Linked List
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* 思路：2中方法：设置一个快走一步的快指针，注意赋值操作顺序。还有一种递归的方法。
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* Tips：递归的方法有点绕，多看下
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*/
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public class lc206 {
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public class ListNode {
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int val;
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ListNode next;
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ListNode(int x) { val = x; }
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}
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public ListNode reverseList(ListNode head) {
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ListNode newHead = null; //头节点变成尾节点，最后要指向null
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while (head != null) {
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ListNode next = head.next;
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head.next = newHead;
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newHead = head;
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head = next;
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}
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return newHead;
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}
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public ListNode reverseList2(ListNode head) { //递归
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return reverseListInt(head, null);
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}
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private ListNode reverseListInt(ListNode head, ListNode newHead) {
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if (head == null)
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return newHead;
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ListNode next = head.next;
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head.next = newHead;
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return reverseListInt(next, head); //尾递归，操作已经完成，最后返回最后结果罢了
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}
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}

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