Leetcode 题解系列（九）

72. Edit Distance

题目要求

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

动态规划

对于一个串a和串b，长度为i和j的子串，那么其最小编辑距离有以下两种情况：
1. a[i - 1] == b[j - 1]
2. a[i - 1] != b[j - 1]
情况一，最小编辑距离等于串a[:i - 1]和串b[ : j - 1]的编辑距离．
情况二，最小编辑距离为串a[:i-1]与串b[:j]，串a[:i]与串b[: j - 1]，串a[:i - 1]与串b[ : j - 1]．

优化

该动态规划计算第i层的时候，只依赖第i - 1层．

class Solution {public:  int minDistance(string word1, string word2) {    vector<int> dp(word2.size() + 1);    for (int i = 0; i <= word2.size(); ++i) {      dp[i] = i;    }    for (int i = 1; i <= word1.size(); ++i) {      int prev = dp[0];      dp[0]++;      for (int j = 1; j <= word2.size(); ++j) {        int dp_store = dp[j];        if (word1[i - 1] == word2[j - 1]) {          dp[j] = prev;        } else {          dp[j] = std::min(std::min(dp[j], dp[j - 1]), prev) + 1;        }        prev = dp_store;      }    }    return dp[word2.size()];  }};