leetcode 712 Minimum ASCII Delete Sum for Two Strings
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Given two strings s1, s2, find the lowest ASCII sum of deleted characters to make two strings equal.
Example 1:
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:
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].
动态规划的经典问题,计算两个字符串之间的编辑距离,只不过这里变成了字符的ASCII码值距离,关键的状态转移方程如下, 时间复杂度O(n^2):
//若相等编辑距离不变if (s1[i-1] == s2[j-1]) { dp[i][j] = dp[i - 1][j - 1];} else {//若不等则有两种情况,取最优的那种情况 dp[i][j] =min(dp[i][j-1]+s2[j-1], dp[i - 1][j ]+s1[i-1]);}
class Solution {public: int minimumDeleteSum(string s1, string s2) { vector<int> t(s2.size()+1); vector<vector<int>> dp; for (int i = 0; i < s1.size()+1; i++) { dp.push_back(t); } for (int i = 0; i < s1.size(); i++) { for (int j = 0; j < s2.size(); j++) { dp[i][j] = INT_MAX; } } dp[0][0] = 0; for (int i = 1; i <= s2.size(); i++) { dp[0][i] = dp[0][i-1]+s2[i-1]; } for (int i = 1; i <= s1.size(); i++) { dp[i][0] =dp[i-1][0]+s1[i-1]; } for (int i = 1; i <= s1.size(); i++) { for (int j = 1; j <= s2.size(); j++) { if (s1[i-1] == s2[j-1]) { dp[i][j] = dp[i - 1][j - 1]; } else { dp[i][j] =min(dp[i][j-1]+s2[j-1], dp[i - 1][j ]+s1[i-1]); } } } return dp[s1.size()][s2.size()]; }};
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