712. Minimum ASCII Delete Sum for Two Strings
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题目描述
解题思路
题目大意如下:要从两个string里删除掉个别字母,使得剩下的字符串相等,并且有个要求——删除掉的字母总的ASCII码值要是最小的。这是典型的动态规划题目,思考之处,我觉得与寻找两个字符串的最长公共子序列有异曲同工之妙,但仔细一想,这两道题目虽然类型相同,但却还不是同一道题目,因为最长公共子序列的数目一定,但却有多种可能,而每种可能对应删除的ASCII码总值却不尽相同,因此还是得老老实实的去做这道题目,不过最长公共子序列的思想却可以在这里借鉴一下:
从两个字符串的开头字符考虑,用二维数组c[i][j]代表字符串s1前i个字符和字符串s2前j个字符实现相同所需要删除的字母对应的ASCII码总值,有三种方法可以到达c[i][j]:
1.c[i-1][j] + s1[i]:由于从c[i-1][j]到c[i][j]是多考虑了s1的一个字符,但是s2字符数没变,所以要想相同,必须删除s1[i],考虑ASCII码总值的话就是加上s1[i]
2.c[i][j-1] + s1[j]:对应于1,这个是多考虑s2的一个字符,所以要删除str2[j]
3.c[i-1][j-1]temp,这里是考虑两个str都加了一个,所以s1[i] =s2[j]时,a=0;s1[i] !=s2[j]时,两个都要删除,即temp=s1[i] +s2[j]
这三种情况每次比较出最小的来,最后返回c[s1.length()][s2.length()]
C++代码实现
int minest(int a, int b, int c) { int min = a; if (b < min) { min = b; } if (c < min) { min = c; } return min;}int minimumDeletSum(string s1, string s2) { int m = s1.length(); int n = s2.length(); int c[m+1][n+1]; c[0][0] = 0; for (int i = 1; i <= m; ++i) { c[i][0] = c[i-1][0] + int(s1[i-1]); } for (int i = 1; i <= n; ++i) { c[0][i] = c[0][i-1] + int(s2[i-1]); } for (int i = 1; i <= m; ++i) { for (int j = 1; j <= n; ++j) { int temp = (s1[i-1] == s2[j-1]) ? 0 : (s1[i-1] + s2[j-1]); c[i][j] = minest(c[i-1][j]+s1[i-1], c[i][j-1]+s2[j-1], c[i-1][j-1]+temp); } } return c[m][n];}
运行结果截图
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