[LeetCode] 712. Minimum ASCII Delete Sum for Two Strings 两个字符串的最小ASCII删除和

2021年09月15日 阅读数:1
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Given two strings s1, s2, find the lowest ASCII sum of deleted characters to make two strings equal.html

Example 1:python

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:post

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:this

  • 0 < s1.length, s2.length <= 1000.
  • All elements of each string will have an ASCII value in [97, 122].

解法:dp,dp[i][j]表示s1的前i个字符和s2的前j个字符所要删除的字符的最小ASCII码和。url

Python:code

# DP with rolling window
class Solution(object):
    def minimumDeleteSum(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: int
        """
        dp = [[0] * (len(s2)+1) for _ in xrange(2)]
        for j in xrange(len(s2)):
            dp[0][j+1] = dp[0][j] + ord(s2[j])

        for i in xrange(len(s1)):
            dp[(i+1)%2][0] = dp[i%2][0] + ord(s1[i])
            for j in xrange(len(s2)):
                if s1[i] == s2[j]:
                    dp[(i+1)%2][j+1] = dp[i%2][j]
                else:
                    dp[(i+1)%2][j+1] = min(dp[i%2][j+1] + ord(s1[i]), \
                                           dp[(i+1)%2][j] + ord(s2[j]))

        return dp[len(s1)%2][-1]

Python:htm

# Time:  O(m * n)
# Space: O(m * n)
class Solution2(object):
    def minimumDeleteSum(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: int
        """
        dp = [[0] * (len(s2)+1) for _ in xrange(len(s1)+1)]
        for i in xrange(len(s1)):
            dp[i+1][0] = dp[i][0] + ord(s1[i])
        for j in xrange(len(s2)):
            dp[0][j+1] = dp[0][j] + ord(s2[j])

        for i in xrange(len(s1)):
            for j in xrange(len(s2)):
                if s1[i] == s2[j]:
                    dp[i+1][j+1] = dp[i][j]
                else:
                    dp[i+1][j+1] = min(dp[i][j+1] + ord(s1[i]), \
                                       dp[i+1][j] + ord(s2[j]))

        return dp[-1][-1]  

C++:blog

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) {
                dp[i][j] = (s1[i - 1] == s2[j - 1]) ? dp[i - 1][j - 1] : min(dp[i - 1][j] + s1[i - 1], dp[i][j - 1] + s2[j - 1]);
            }
        }
        return dp[m][n];
    }
};

C++:element

class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        int m = s1.size(), n = s2.size();
        vector<int> dp(n + 1, 0);
        for (int j = 1; j <= n; ++j) dp[j] = dp[j - 1] + s2[j - 1];
        for (int i = 1; i <= m; ++i) {
            int t1 = dp[0];
            dp[0] += s1[i - 1];
            for (int j = 1; j <= n; ++j) {
                int t2 = dp[j];
                dp[j] = (s1[i - 1] == s2[j - 1]) ? t1 : min(dp[j] + s1[i - 1], dp[j - 1] + s2[j - 1]);
                t1 = t2;
            }
        }
        return dp[n];
    }
};

C++:字符串

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 i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (s1[i - 1] == s2[j - 1]) dp[i][j] = dp[i - 1][j - 1] + s1[i - 1];
                else dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        int sum1 = accumulate(s1.begin(), s1.end(), 0);
        int sum2 = accumulate(s2.begin(), s2.end(), 0);
        return sum1 + sum2 - 2 * dp[m][n];
    }
};

  

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