zoukankan      html  css  js  c++  java
  • LeetCode 712. Minimum ASCII Delete Sum for Two Strings

    原题链接在这里:https://leetcode.com/problems/minimum-ascii-delete-sum-for-two-strings/

    题目:

    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].

    题解:

    For two strings, dp[i][j] denotes s1 up to i and s2 up to j, minimum ASCII delete sum for two substrings.

    There could be 2 cases:

    case 1:

    s1.charAt(i) == s2.charAt(j), then it doesn't need to delete last characters, dp[i][j] = dp[i-1][j-1].

    case 2:

    s1.charAt(i) != s2.charAt(j). then minimum delete could comes from 2 possibilities, take the minimum:

    • delete s1.charAt(i). dp[i][j] = dp[i-1][j] + s1.charAt(i).
    • delete s2.charAt(j). dp[i][j] = dp[i][j-1] + s2.charAt(j).

    Time Complexity: O(m*n). m = s1.length(). n = s2.length().

    Space: O(m*n).

    AC Java:

     1 class Solution {
     2     public int minimumDeleteSum(String s1, String s2) {
     3         int m = s1.length();
     4         int n = s2.length();
     5         int [][] dp = new int[m+1][n+1];
     6         
     7         for(int i = 1; i<=m; i++){
     8             dp[i][0] = dp[i-1][0]+s1.charAt(i-1);
     9         }
    10         
    11         for(int j = 1; j<=n; j++){
    12             dp[0][j] = dp[0][j-1] + s2.charAt(j-1);
    13         }
    14         
    15         for(int i = 1; i<=m; i++){
    16             for(int j = 1; j<=n; j++){
    17                 if(s1.charAt(i-1) == s2.charAt(j-1)){
    18                     dp[i][j] = dp[i-1][j-1];
    19                 }else{
    20                     dp[i][j] = Math.min(dp[i-1][j]+s1.charAt(i-1), dp[i][j-1]+s2.charAt(j-1));
    21                 }
    22             }
    23         }
    24         
    25         return dp[m][n];
    26     }
    27 }

    类似Longest Common SubsequenceDelete Operation for Two StringsEdit Distance.

  • 相关阅读:
    介绍Shiro Annotation及Shiro标签的用法
    SpringMVC+Apache Shiro+JPA(hibernate)案例教学(四)基于Shiro验证用户权限,且给用户授权
    SpringMVC+Apache Shiro+JPA(hibernate)案例教学(一)整合配置
    Spring MVC之@RequestMapping 详解
    vimrc
    sk_buff深度解析
    ieee80211ax简介
    新基地
    dos格式迭代转为unix
    ath10k MAC地址
  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11450840.html
Copyright © 2011-2022 走看看