第十七周LeetCode算法题

题目名称：72. Edit Distance

题目难度：Hard

题目描述：

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

题目分析：

题目要求找出两个字符串之间的“编辑距离”，即用最小的步骤将两个字符串变成一样的。
可以用动态规划思想去做。
设需要计算编辑距离的两个字符串分别为x[1..m]和 y[1..n],维护一个二维数组E[i][j],表示a[1..i]和b[1..j]的编辑距离.
E[i][j]与前面的值有关
要么就是在a[1..i-1]和b[1..j]的编辑距离的基础上，a字符再向前一步，变成a[1..i]和b[1..j]，那么此时E[i][j]就是E[i - 1][j] + 1
要么就是在a[1..i]和b[1..j-1]的编辑距离的基础上，b字串再向前一步，变成a[1..i]和b[1..j]，那么此时E[i][j]就是E[i][j - 1] + 1
要么就是在a[1..i-1]和b[1..j-1]的编辑距离的基础上，a、b的字符均再向前一步，变成a[1..i]和b[1..j]，那么此时E[i][j]就是E[i - 1][j - 1] + diff(x[i - 1], y[j - 1]
其中，当x[i]=y[j]时， diff(i, j)=0;否则diff(i, j)=1.

即：
当i = 0时，E[i][j] = j;当j = 0时，E[i][j] = i;
当i > 0,j > 0时，E[i][j] = min(E[i - 1][j] + 1, E[i][j - 1] + 1, E[i - 1][j - 1] + diff(x[i - 1], y[j - 1]))
其中，当x[i]=y[j]时， diff(i, j)=0;否则diff(i, j)=1.

最后AC的代码是：

class Solution {public:    bool diff(char a, char b) {        return !(a == b);    }    int min(int a, int b, int c) {        if (a <= b && a <= c) return a;        if (b <= a && b <= c) return b;        if (c <= a && c <= b) return c;    }    int minDistance(string word1, string word2) {        if (word1 == "") return word2.size();        if (word2 == "") return word1.size();        if (word1 == word2) return 0;        int E[word1.size() + 1][word2.size() + 1];        for (int i = 0; i <= word1.size(); ++i) {            E[i][0] = i;        }        for (int i = 0; i <= word2.size(); ++i) {            E[0][i] = i;        }        for (int i = 1; i <= word1.size(); ++i) {            for (int j = 1; j <= word2.size(); ++j) {                E[i][j] = min(E[i - 1][j] + 1, E[i][j - 1] + 1, E[i - 1][j - 1] + diff(word1[i - 1], word2[j - 1]));            }        }        return E[word1.size()][word2.size()];    }};