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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: （Hard）
a) Insert a character
b) Delete a character
c) Replace a character
分析：
双序列动态规划问题，dp[i][j]表示第一个序列i ...和 第二个序列j ...；
对应本题：
1. 状态： dp[i][j]表示第一个字符串前i个字符到第二个字符串前j个字符需要的编辑距离；
2. 递推关系：
如果 s1[i] == s2[j]
dp[i][j] = min(dp[i-1][j-1] //表示修改
,dp[i-1][j] + 1 //表示删除
,dp[i][j-1] + 1) //表示增加
如果 s1[i] != s2[j]
dp[i][j] = min(dp[i-1][j-1] ,dp[i-1][j],dp[i][j-1]) + 1
3. 初始化：
dp[i][0] = i; i = 1...m;
dp[0][j] = j; j = 1...n;
代码：
1 class Solution { 2 public: 3 int minDistance(string word1, string word2) { 4 //dp[i][j]表示第一个字符串前i个字符到第二个字符串前j个字符需要的编辑距离； 5 int m = word1.size(), n = word2.size() ; 6 int length = max(m,n); 7 int dp[length + 1][length + 1]; 8 dp[0][0] = 0; 9 for(int i = 1;i <= m;++i){ 10 dp[i][0] = i; 11 } 12 for(int i = 1; i <= n;++i){ 13 dp[0][i] = i; 14 } 15 for(int i = 1; i <= m; ++i){ 16 for(int j = 1; j <= n; ++j){ 17 if(word1[i - 1] != word2[j - 1]){ 18 dp[i][j] = min( min(dp[i-1][j-1],dp[i-1][j]), dp[i][j-1]) + 1; 19 } 20 else{ 21 dp[i][j] = min( min(dp[i-1][j-1], dp[i-1][j] + 1), dp[i][j-1] + 1); 22 } 23 } 24 } 25 return dp[m][n]; 26 } 27 };
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原文地址：http://www.cnblogs.com/wangxiaobao/p/5901877.html
