Edit Distance
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
思路：典型的动态规划，dp[i][j]表示word1和word2的编辑距离，当word1[i] == word2[j]时，dp[i][j] == dp[i-1][j-1]，当word1[i] ！= word2[j]时，
dp[i][j] = min(dp[i-1][j-1],min(dp[i-1][j],dp[i][j-1]))+1
class Solution {
public:
int minDistance(string word1, string word2) {
int length1 = word1.size(),length2 = word2.size(),i,j;
vector
> dp(length1+1);
for(i = 0;i <= length1;++i)
{
vector
tmp(length2+1,0); dp[i] = tmp; } for(i = 1; i <= length1;++i)dp[i][0] = i; for(j = 1; j <= length2;++j)dp[0][j] = j; for(i = 1; i <= length1;++i) { for(j = 1;j <= length2;++j) { 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[length1][length2]; } };