Interleaving String

 

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.

dp[i j] 表示s1[0 - i]是否与s2[0 - j]是交叉串。

• 则空串与空串为true,即dp[0][0] = 1；
• dp[i][j] = (dp[i][j - 1] && s2[j - 1] == s3[i + j - 1]) || (dp[i - 1][j] && s1[i - 1] == s3[i + j - 1]);
   dp[i][j]为真时，肯定是s1[0,i]与s2[0-j]交叉组成 串s3 的 [0 ,i + j -1]，s3[i + j - 1] = s1[i] OR s2[j],

 则 当s3[i + j - 1] == s1[i] && dp[i][j - 1] && s2[j - 1]  或者dp[i - 1][j] && s1[i - 1] == s3[i + j - 1]);时为真

 

• 这里要注意当i 或者j 为0时 i-1 (j - 1)越界的问题！

 

• 对于两个数组i,j关系的问题，考虑局部0-i 和0-j,最后以dp公式，得出0-i 与0-j的关系即可！

 

 

 

 

 

class Solution {public:    bool isInterleave(string s1, string s2, string s3) {int len1 = s1.size();int len2 = s2.size();int len3 = s3.size();//长度不相等，falseif(len1 + len2 != len3)return false;vector<vector<int> >dp;vector<int>v;for(int i = 0; i <= len1; i++){v.clear();for(int j = 0; j <= len2; j++)v.push_back(0);dp.push_back(v);}//在对字符串，数组固定取值时，先判断下标是否越界if(len1 && s1[0] == s3[0])dp[1][0] = 1;if(len2 && s2[0] == s3[0])dp[0][1] = 1;  dp[0][0] = 1;for(int i = 0; i <= len1; i++){for(int j = 0; j <= len2; j++){//对于可能越界的直接用if 省事if(!i && j)//i = 0时只需考虑dp[i][j - 1]  、s2[j - 1]dp[i][j] = dp[i][j - 1] && s2[j - 1] == s3[i + j - 1];else if(i && !j)//i = 0时只需考虑dp[i - 1][j]  、s2[i - 1]，即dp[0][i]是否是交叉串dp[i][j] = dp[i - 1][j] && s1[i - 1] == s3[i + j - 1];else if(i && j)dp[i][j] = (dp[i][j - 1] && s2[j - 1] == s3[i + j - 1]) || (dp[i - 1][j] && s1[i - 1] == s3[i + j - 1]);}}return dp[len1][len2];    }};