LeetCode72 Edit Distance(编辑距离)

本题是LeetCode 第72题，可以用动态规划来求解。github地址点击链接。

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 operations permitted on a word:

• Insert a character
• Delete a character
• Replace a character

---

Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation: 
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

分析

动态规划的典型题：求最小步长，可以通过最优子结构n-1来推导n的情况
状态转移方程：
dp二维数组，用来保存word1 第0...i个字符变化到word2 第0...j个字符的最小步长
一共有三种策略：删，添，改。
1）若 word1[i] == word2[j] ，则不用进行操作dp[i+1][j+1] = dp[i][j]
2) 若 word1[i] != word2[j] , 则可以通过三种策略：删，添，改当前字母，
使word1[0...i]->word2[0...j]字符串。三种策略对应如下：
    删：dp[i+1][j+1] = dp[i][j+1] + 1
    添：dp[i+1][j+1] = dp[i+1][j] + 1
    改：dp[i+1][j+1] = dp[i][j] +1
取出最小更新到dp[i+1][j+1]就是状态转移方程，即dp[i+1][j+1] = min(dp[i+1][j],min(dp[i][j],dp[i][j+1]))+1;

例

  word1="horse" word2="ros"
  dp[][]  "" r o s
        "" 0 1 2 3
        h  1 1 2 3 
        o  2 2 1 2
        r  3 2 2 2
        s  4 3 3 2
        e  5 4 4 3

注：由于本例dp数组只需要记录当前行以及上一行（i和i-1行），其他行数都不需要，可以通过3行的滚动数组来存储，用来降低空间复杂度。

#include <vector>
#include <string>
#include <algorithm>
using namespace std;

class Solution {
public:
    int minDistance(string word1, string word2) {

        int len1 = word1.size();
        int len2 = word2.size();

        //若为空，则特殊处理
        if(!len1) return len2; 
        if(!len2) return len1;

        //分配dp二维数组，用来保存word1 第0...i个字符变化到word2 第0...j个字符的最小步长
        vector<vector<int>> dp(len1+1,vector<int>(len2+1));
        for(int i=0; i <= len2; ++i)
            dp[0][i] = i;
        for(int i=0; i <= len1; ++i)
            dp[i][0] = i;

        //利用状态转移方程，填充数组
        for(int i=0; i<len1; ++i)
        {
            for(int j=0; j<len2; ++j)
            {
                if( word1[i] == word2[j])
                    dp[i+1][j+1] = dp[i][j];
                else
                {
                    dp[i+1][j+1] = min(dp[i+1][j],min(dp[i][j],dp[i][j+1]))+1;
                }
            }
        }
        return dp[len1][len2];
    }
};