97/100 Interleaving String/Same Tree

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

For example,
Given:
s1 = “aabcc”,
s2 = “dbbca”,

When s3 = “aadbbcbcac”, return true.
When s3 = “aadbbbaccc”, return false.

可以用递归做，每匹配s1或者s2中任意一个就递归下去。但是会超时。

因此考虑用动态规划做。

s1, s2只有两个字符串，因此可以展平为一个二维地图，判断是否能从左上角走到右下角。

当s1到达第i个元素，s2到达第j个元素:

地图上往右一步就是s2[j-1]匹配s3[i+j-1]。

地图上往下一步就是s1[i-1]匹配s3[i+j-1]。

示例：s1=”aa”,s2=”ab”,s3=”aaba”。标1的为可行。最终返回右下角。

    0  a  b0   1  1  0a   1  1  1a   1  0  1

class Solution {public:    bool isInterleave(string s1, string s2, string s3) {        int m = s1.size();        int n = s2.size();        if(m+n != s3.size())            return false;        vector<vector<bool> > path(m+1, vector<bool>(n+1, false));        for(int i = 0; i < m+1; i ++)        {            for(int j = 0; j < n+1; j ++)            {                if(i == 0 && j == 0)                // start                    path[i][j] = true;                else if(i == 0)                    path[i][j] = path[i][j-1] & (s2[j-1]==s3[j-1]);                else if(j == 0)                    path[i][j] = path[i-1][j] & (s1[i-1]==s3[i-1]);                else                    path[i][j] = (path[i][j-1] & (s2[j-1]==s3[i+j-1])) || (path[i-1][j] & (s1[i-1]==s3[i+j-1]));            }        }        return path[m][n];    }};

笔记：字符串处理，直观可以想到用递归做的题，基本都可以用DP求解.100 Same Tree:

Given two binary trees, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical and the nodes have the same value.

Example 1:

Input:     1         1          / \       / \         2   3     2   3        [1,2,3],   [1,2,3]Output: true

Example 2:

Input:     1         1          /           \         2             2        [1,2],     [1,null,2]Output: false

Example 3:

Input:     1         1          / \       / \         2   1     1   2        [1,2,1],   [1,1,2]Output: false

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool isSameTree(TreeNode* p, TreeNode* q) {        if(p == NULL && q == NULL)            return true;        else if(p == NULL && q != NULL)            return false;        else if(p != NULL && q == NULL)            return false;        else if(p->val != q->val)            return false;        else            return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);     }};

笔记：简单的递归调用，思路要清晰。