首页 > 其他 > 详细

72. Edit Distance

时间:2017-05-15 09:04:16      阅读:196      评论:0      收藏:0      [点我收藏+]

Problem statement:

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

Solution:

This is also a DP problem, which is a little complicated than 583. Delete Operation for Two Strings. We can do two more operations: insert and replace.

The dp formular is complicated when word1[i] != word2[j].

But the initialization and return value is the same.

class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.size();
        int n = word2.size();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        
        // initialize
        for(int i = 0; i <= m; i++){
            dp[i][0] = i;
        }
        for(int j = 0; j <= n; j++){
            dp[0][j] = j;
        }
        // dynamic programming
        for(int i = 1; i <= m; i++){
            for(int j = 1; j <= n; j++){
                // two character is equal
                if(word1[i - 1] == word2[j - 1]){
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;
                }
            }
        }
        return dp[m][n];
    }
};

72. Edit Distance

原文:http://www.cnblogs.com/wdw828/p/6854765.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!