description:

my solution:

class Solution {
public:
	int minimumDeleteSum(string s1, string s2) {
		//i is the length of s1
		//j is the length of s2
		//dp[i][j] = min(dp[i-1][j] + s1[i], dp[i][j-1] + s2[j])
		int m = s1.length(), n = s2.length();
		vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
		for (int j = 1; j < n + 1; j++) dp[0][j] = dp[0][j-1] + s2[j-1];
		for (int i = 1; i < m + 1; i++) {
			dp[i][0] = dp[i - 1][0] + s1[i-1];
			for (int j = 1; j < n + 1; j++) {
				if (s1[i - 1] == s2[j - 1])
					dp[i][j] = dp[i - 1][j - 1];
				else
					dp[i][j] = min(dp[i - 1][j] + s1[i - 1], dp[i][j - 1] + s2[j - 1]);
			}
		}
		return dp[m][n];
	}
};

thought:

利用动态规划求解，算是比较简单的题目了，逐渐分别增长s1和s2的长度。 别人给出了占用更少空间和长度的算法。

Better ways:

by zestypanda

class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        int m = s1.size(), n = s2.size();
        vector<int> dp(n+1, 0);
        for (int j = 1; j <= n; j++)
            dp[j] = dp[j-1]+s2[j-1];
        for (int i = 1; i <= m; i++) {
            int t1 = dp[0];
            dp[0] += s1[i-1];
            for (int j = 1; j <= n; j++) {
                int t2 = dp[j];
                dp[j] = s1[i-1] == s2[j-1]? t1:min(dp[j]+s1[i-1], dp[j-1]+s2[j-1]);
                t1 = t2;
            }
        }
        return dp[n];
    }
};