leetcode-712

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字数统计: 415

712 Minimum ASCII Delete Sum for Two Strings

Given two strings s1, s2, find the lowest ASCII sum of deleted characters to make two strings equal.

Example 1:

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Input: s1 = "sea", s2 = "eat"
Output: 231
Explanation: Deleting "s" from "sea" adds the ASCII value of "s" (115) to the sum.
Deleting "t" from "eat" adds 116 to the sum.
At the end, both strings are equal, and 115 + 116 = 231 is the minimum sum possible to achieve this.

Example 2:

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Input: s1 = "delete", s2 = "leet"
Output: 403
Explanation: Deleting "dee" from "delete" to turn the string into "let",
adds 100[d]+101[e]+101[e] to the sum.  Deleting "e" from "leet" adds 101[e] to the sum.
At the end, both strings are equal to "let", and the answer is 100+101+101+101 = 403.
If instead we turned both strings into "lee" or "eet", we would get answers of 433 or 417, which are higher.

Note:

0 < s1.length, s2.length <= 1000.

All elements of each string will have an ASCII value in [97, 122].

这个就是动态规划求解，所以要先建立dp[][]二维数组，之后建立状态方程，需要循环遍历，依次从0开始查找dp[i][j]的最小值。代码如下

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class Solution {
    public int minimumDeleteSum(String s1, String s2) {
        int[][] dp = new int[s1.length()+1][s2.length()+1];   
        for (int i = 1; i < s1.length()+1; i++) {  
            dp[i][0] = dp[i-1][0] + s1.charAt(i-1);   
        }
        for (int j = 1; j < s2.length()+1; j++) {
            dp[0][j] = dp[0][j-1] + s2.charAt(j-1);
        }
    for (int i = 1; i <= s1.length(); i++) {
        for (int j = 1; j <= s2.length(); j++) {
            if(s1.charAt(i-1) == s2.charAt(j-1))  // 判断 s1和s2的字符是否相等 
            {
                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[s1.length()][s2.length()];
}
}
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