72. Edit Distance

原网址为https://leetcode.com/problems/edit-distance/description/

一.题目

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

二、解法

class Solution {public:    int minDistance(string word1, string word2) {        int len1 = word1.size() + 1;        int len2 = word2.size() + 1;        vector<vector<int> >edit(len1);        for (int i = 0; i < len1; i++) {            edit[i] = vector<int>(len2);        }                for (int j = 0; j < len2; j++) {            edit[0][j] = j;        }                for (int i = 0; i < len1; i++) {            edit[i][0] = i;        }                for (int i = 1; i < len1; i++) {            for (int j = 1; j < len2; j++) {                edit[i][j] = min(edit[i - 1][j] + 1, edit[i][j - 1] + 1, edit[i - 1][j - 1] + plusOneIfNotEqual(word1[i - 1], word2[j - 1]));            }        }        return edit[len1 - 1][len2 - 1];    }        int min(int a, int b, int c) {        int m = a;        m = m < b ? m : b;        m = m < c ? m : c;        return m;    }        int plusOneIfNotEqual(char a, char b) {        return (a == b ? 0 : 1);    }};

这可以用动态规划来解决。

edit[i][j]表示字符串word1转换为字符串word2的edit distance，那么

当i == 0时， edit[0][j] = j

当j == 0 时， edit[i][0] = i

当i >= 1 && j >= 1时， edit[i][j] = min(edit[i - 1][j] + 1, edit[i][j - 1] + 1, edit[i - 1][j - 1] + plusOneIfNotEqual(word1[i - 1], word2[j - 1]));

min表示三个数的最小值，plusOneIfNotEqual表示如果输入a和b相等，就返回1，不等为0