583. Delete Operation for Two Strings

标签：ring   problems   int   ems   ndis   case   sub   max   NPU   

Given two words word1 and word2, find the minimum number of steps required to make word1 and word2 the same, where in each step you can delete one character in either string.

Example 1:

Input: "sea", "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".

Note:

1. The length of given words won‘t exceed 500.
2. Characters in given words can only be lower-case letters.

Approach #1: DP. [Java]

class Solution {
    public int minDistance(String word1, String word2) {
        int len1 = word1.length();
        int len2 = word2.length();
        int[][] dp = new int[len1+1][len2+1];
        for (int i = 0; i <= len1; ++i) {
            for (int j = 0; j <= len2; ++j) {
                if (i == 0 || j == 0) {
                    dp[i][j] = 0;
                    continue;
                }
                dp[i][j] = word1.charAt(i-1) == word2.charAt(j-1) ? dp[i-1][j-1] + 1 : Math.max(dp[i][j-1], dp[i-1][j]);
            }
        }
        int val = dp[len1][len2];
        return len1 - val + len2 - val;
    }
}

Analysis:

To make them identical, just find the longest common subsequence. The rest of the characters have to be deleted from the both the strings. which does not belong to longest common subsquence.

Reference:

https://leetcode.com/problems/delete-operation-for-two-strings/discuss/103214/Java-DP-Solution-(Longest-Common-Subsequence)

583. Delete Operation for Two Strings

标签：ring   problems   int   ems   ndis   case   sub   max   NPU   

原文地址：https://www.cnblogs.com/ruruozhenhao/p/10755623.html