leetcode712-两个字符串的最小ASCII删除和

两个字符串的最小ASCII删除和

• dp

dp[i][j]代表s1[0:i]和s2[0:j]两个字符串的最小删除值。转移方程如下：

1. 如果i == 0且j == 0，则dp[i][j] = 0
2. 如果i == 0或j == 0，则dp[i][j] = dp[i-1][j]+s1.charAt(i-1)或dp[i][j] = dp[i][j-1]+s2.charAt(j-1)
3. 除此之外，需要判断两个字符是否相等
   • 如果相等，则dp[i][j] = dp[i-1][j-1]
   • 如果不相等，则计算字符和对应的转移状态之和，挑选较小的一个，dp[i][j] = Math.min(dp[i-1][j]+s1.charAt(i-1), dp[i][j-1]+s2.charAt(j-1))

class Solution {
    public int minimumDeleteSum(String s1, String s2) {
        int m = s1.length(), n = s2.length(), dp[][] = new int[m+1][n+1];
        for(int i = 0; i <= m; i++){
            for(int j = 0; j <= n; j++){
                if(i == 0 && j == 0)    dp[i][j] = 0;
                else if(i == 0) dp[i][j] = dp[i][j-1]+s2.charAt(j-1);
                else if(j == 0) dp[i][j] = dp[i-1][j]+s1.charAt(i-1);
                else{
                    if(s1.charAt(i-1) == s2.charAt(j-1))    dp[i][j] = dp[i-1][j-1];
                    else    dp[i][j] = Math.min(dp[i-1][j]+s1.charAt(i-1), dp[i][j-1]+s2.charAt(j-1));
                }
            }
        }
        return dp[m][n];
    }
}