LeetCode 712. Minimum ASCII Delete Sum for Two Strings [Medium]
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原题地址
题目内容
题目分析
标准的动态规划问题,题目的意思为删除s1,s2中的字符,使得最后s1=s2,要求使得删除的字符的值的和最小。dp[i][j]为字符串1前i个字符与字符串2前j个字符相等,需要减去的字符值。当s1[i-1]=s2[j-1]时,dp[i][j]=dp[i-1][j-1].当s1[i-1]!=s2[j-1]时,有两种选择,一删除s1的字符,那么dp[i][j]=dp[i-1][j]+s1[i-1],二是删除s2的字符,dp[i][j]=dp[i][j-1]+s2[j-1],dp[i][j]显然是选两者当中代价最小的。
代码分析
class Solution {public: int minimumDeleteSum(string s1, string s2) { int n = s1.size(); int m = s2.size(); int dp[n+1][m+1] = {0}; for(int i = 1; i <= m; i++){ dp[0][i] = dp[0][i-1]+s2[i-1]; } for(int i = 1; i <= n; i++){ dp[i][0] = dp[i-1][0] + s1[i-1]; for(int j = 1; j <= m; 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[n][m]; }};
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