LeetCode:Minimum ASCII Delete Sum for Two Strings

题目链接：https://leetcode.com/problems/minimum-ascii-delete-sum-for-two-strings/description/

题目介绍：给出2个字符串，找出删除掉ASCII码值和最小的字符使得2个字符串相同。

解题思路：设需要编辑的2个字符串为s1[1...m]和s2[1...n]，用S(i,j)表示s1[1...i]和s2[1...j]的删除字符的最小ASCII码值之和。

S(0,0)=0;

当i=0时，S(i,j)=s2[j]+S(i,j-1);

当j=0时，S(i,j)=s1[i]+S(i-1,j);

当i>0且j>0时，S(i,j)=min{s1[i]+S(i-1,j), s2[j]+S(i,j-1), cost(i,j)+S(i-1,j-1)}

其中，当s1[i]=s2[j]时，cost(i,j)=0;否则cost(i,j)=s1[i]+s2[j]。

代码如下：

class Solution {public:    int minimumDeleteSum(string s1, string s2) {    int size1 = s1.size(), size2 = s2.size();    if (size1 == 0 && size2 == 0) {        return 0;    }    else if (size1 == 0) {        int sum = 0;        for (int i = 0; i < size2; i++) {            sum += s2[i];        }        return sum;    }    else if (size2 == 0) {        int sum = 0;        for (int i = 0; i < size1; i++) {            sum += s1[i];        }        return sum;    }    else {        int** S = new int*[size1+1];        for (int i = 0; i < size1+1; i++) {            S[i] = new int[size2+1];        }        S[0][0] = 0;        for (int i = 1; i < size1 + 1; i++) {            S[i][0] = S[i-1][0] + s1[i-1];        }        for (int i = 1; i < size2 + 1; i++) {            S[0][i] = S[0][i-1] + s2[i-1];        }        for (int i = 1; i < size1 + 1; i++) {            for (int j = 1; j < size2 + 1; j++) {                int d1 = s1[i-1] + S[i - 1][j];                int d2 = s2[j-1] + S[i][j - 1];                int d0 = S[i - 1][j - 1];                if (s1[i-1] != s2[j-1]) d0 += s1[i-1] + s2[j-1];                S[i][j] = min(d0, d1, d2);            }        }        return S[size1][size2];    }}int min(int a, int b, int c) {    int min = (a < b) ? a : b;    return (min < c) ? min : c;}};