标签：mini eps can turn style lcs sha title cte
题目：
Given two words word1 and word2, find the minimum number of steps required to make word1and word2 the same, where in each step you can delete one character in either string.
Example 1:
Input: "sea", "eat" Output: 2 Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
Note:
思路：
这题属于动态规划，而且是字符串，就想到lcs，读完题发现就是一个fcs，只需要长度，不需要具体串。
通过操作使最后字符串一样，求操作次数，即删除多少字符。
求出最长公共子序列后，长度之和减去两倍的最长公共子序列长度，得结果。
代码：
1 class Solution { 2 public: 3 int minDistance(string word1, string word2) { 4 int l1 = word1.length(); 5 int l2 = word2.length(); 6 int m[501][501] = {0}; 7 for(int i = 1 ;i <= l1;i++){ 8 for(int j = 1; j <= l2;j++){ 9 m[i][j] = max(m[i-1][j],m[i][j-1]); 10 if(word1[i-1] == word2[j-1]){ 11 m[i][j] = max(m[i][j],m[i-1][j-1]+1); 12 } 13 } 14 } 15 return l1 + l2 - 2 * m[l1][l2]; 16 } 17 };
leetcode delete-operation-for-two-strings
标签：mini eps can turn style lcs sha title cte
原文地址：https://www.cnblogs.com/zhibin123/p/10364501.html
