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@ -96,6 +96,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
| 94 |[Binary Tree Inorder Traversal](https://leetcode.com/problems/binary-tree-inorder-traversal/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/94.%20Binary%20Tree%20Inorder%20Traversal.md)|Medium| |
| 95 |[Unique Binary Search Trees II](https://leetcode.com/problems/unique-binary-search-trees-ii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/95.%20Unique%20Binary%20Search%20Trees%20II.md)|Medium| |
| 96 |[Unique Binary Search Trees](https://leetcode.com/problems/unique-binary-search-trees/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/96.%20Unique%20Binary%20Search%20Trees.md)|Medium| |
| 97 |[Interleaving String](https://leetcode.cn/problems/interleaving-string/)|[C++](solutions/97.%20Interleaving%20String.md)|Medium| |
| 98 |[Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/98.%20Validate%20Binary%20Search%20Tree.md)|Medium| |
| 100 |[Same Tree](https://leetcode.com/problems/same-tree)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/100.%20Same%20Tree.md)|Easy| |
| 101 |[Symmetric Tree](https://leetcode.com/problems/symmetric-tree)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/101.%20Symmetric%20Tree.md)|Easy| |

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solutions/97. Interleaving String.md Normal file
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# [97. Interleaving String](https://leetcode.cn/problems/interleaving-string/)
# 思路
仔细分析问题不难发现会涉及到大量子问题，所以我们考虑用动态规划求解。我们定义 dp[i][j] 表示 s1 的前 i 个元素和 s2 的前 j 个元素是否能交错组成 s3 的前 i + j 个元素。可得到初始状态和状态转移方程为：
```
初始态（即都为空串）：dp[0][0] = true
转移方程：dp[i][j] = (dp[i-1][j] && s3[i+j-1] == s1[i-1]) || (dp[i][j-1] && s3[i+j-1] == s2[j-1])
```
再考虑下边界条件就不难写出代码。另外由于 dp[i][j] 只和左上元素有关，故可以考虑使用滚动数组优化空间复杂度为O(s2.size())。
# C++
``` C++
class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
if(s1.size() + s2.size() != s3.size()) return false;
int n1 = s1.size(), n2 = s2.size(), n3 = s3.size();
if(n1 == 0) return s2 == s3;
if(n2 == 0) return s1 == s3;
// vector<vector<bool>>dp(n1 + 1, vector<bool>(n2 + 1, false));
// for(int i = 0; i <= n1; i++){
// for(int j = 0; j <= n2; j++){
// if(i == 0 && j == 0) dp[i][j] = true;
// else if(i > 0 && dp[i-1][j] && s3[i+j-1] == s1[i-1]) dp[i][j] = true;
// else if(j > 0 && dp[i][j-1] && s3[i+j-1] == s2[j-1]) dp[i][j] = true;
// // else dp[i][j] = false;
// }
// }
// return dp[n1][n2];
// 滚动数组空间优化
vector<bool>dp(n2 + 1, false);
for(int i = 0; i <= n1; i++){
for(int j = 0; j <= n2; j++){
if(i == 0 && j == 0) dp[j] = true;
else if(i > 0 && dp[j] && s3[i+j-1] == s1[i-1]) dp[j] = true;
else if(j > 0 && dp[j-1] && s3[i+j-1] == s2[j-1]) dp[j] = true;
else dp[j] = false; // 注意这里需重置成false, 因为dp[i-1][j]可能等于true
}
}
return dp[n2];
}
};
```