第18周Minimum ASCII Delete Sum for Two Strings

Minimum ASCII Delete Sum for Two Strings

Leetcode algorithms problem 712：Minimum ASCII Delete Sum for Two Strings

• 問題描述

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

• 例子

  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.

• 思路

  設dp [i] [j]是s1.substr（0，i）和s2.substr（0，j）的成本。
  Base case: dp[0][0] = 0
  target: dp[m][n]
  if s1[i-1] = s2[j-1] // 不需要刪除
  dp[i][j] = dp[i-1][j-1];
  else // 刪除s1[i-1]或s2[j-1]
  dp[i][j] = min(dp[i-1][j]+s1[i-1], dp[i][j-1]+s2[j-1]);

代碼

class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        int m = s1.size(), n = s2.size();
        vector<vector<int>> dp(m+1, vector<int>(n+1, 0));
        for (int j = 1; j <= n; j++)
            dp[0][j] = dp[0][j-1]+s2[j-1];
        for (int i = 1; i <= m; i++) {
            dp[i][0] = dp[i-1][0]+s1[i-1];
            for (int j = 1; j <= n; j++) {
                if (s1[i-1] == s2[j-1])
                    dp[i][j] = dp[i-1][j-1];
                else 
                    dp[i][j] = min(dp[i-1][j]+s1[i-1], dp[i][j-1]+s2[j-1]);
            }
        }
        return dp[m][n];
    }
};

時間複雜度: O(nm)
空間複雜度: O(nm)

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