20190127

Author: mJackie

code/lc108.java

@@ -0,0 +1,29 @@
+package code;
+/*
+ * 108. Convert Sorted Array to Binary Search Tree
+ * 题意：将有序数组转换为二叉平衡树
+ * 难度：Easy
+ * 分类：Tree, Depth-first Search
+ * 思路：类似二分查找，每次把数组劈成两半
+ * Tips：Bingo!
+ */
+public class lc108 {
+    public class TreeNode {
+        int val;
+        TreeNode left;
+        TreeNode right;
+        TreeNode(int x) { val = x; }
+    }
+    public TreeNode sortedArrayToBST(int[] nums) {
+        TreeNode tn = helper(nums, 0, nums.length-1);
+        return tn;
+    }
+    public TreeNode helper(int[] nums, int left, int right){
+        if(left<right) return null;
+        int mid = (left+right)/2;
+        TreeNode tn = new TreeNode(nums[mid]);
+        tn.left = helper(nums, left, mid-1);
+        tn.right = helper(nums, mid+1, right);
+        return tn;
+    }
+}

code/lc91.java

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+package code;
+/*
+ * 91. Decode Ways
+ * 题意：1~26代表了26个字母
+ * 难度：Medium
+ * 分类：String, Dynamic Programming
+ * 思路：动态规划，和上台阶问题很像
+ * Tips：Bingo!
+ */
+public class lc91 {
+    public static void main(String[] args) {
+        System.out.println(numDecodings("99"));
+    }
+    public static int numDecodings(String s) {
+        if(s.length()==0) return 0;
+        int[] dp = new int[s.length()+1];
+        dp[0] = 1;
+        int ch = s.charAt(0)-'0';
+        if( ch<1 || ch>9 ) return 0;    //判断第一个字符是否符合规范
+        else dp[1] = 1;
+        for (int i = 2; i < s.length()+1 ; i++) {
+            int a = s.charAt(i-2)-'0';
+            int b = s.charAt(i-1)-'0';
+            if( (a*10+b)<=26 && (a*10+b)>=10 )  //可能存在两个字符10符合，但一个字符0不符合的情况，所以+=
+                dp[i] += dp[i-2];
+            if( b>0 && b<=9 )
+                dp[i] += dp[i-1];
+            if(dp[i]==0) return 0;  //解析错误
+        }
+        return dp[s.length()];
+    }
+}