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  b

0   1  1  0

a   1  1  1

a   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]
;
}
};

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。