[LeetCode] DP 之 Edit Distance
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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:
a) Insert a character
b) Delete a character
c) Replace a character
分析
这题就是课本上的编辑距离可以分为3种情况来分析
对于两个字符串s1, s2 , e[i][j]为 s1[0….i-1], s2[0….j-1] 的最小编辑距离,那么对于e[i][j],假设我们已经知道了其子问题的所有解,而且s1[i-1] != s1[j-1]那么我们有三种可以选择的处理方式
- 插入。对于e[i][j-1] 插入字符s1[i-1] 使得最右端的字符相等
- 删除。对于e[i][j] 删除字符s1[i-1] 使得最右端字符相等
- 替换。对于e[i][j],用s2[j-1] 替换s1[i-1]
因此表达式为: e[i][j] = min(1 + e[i-1][j], min(1+ e[i][j-1] , diff + e[i-1][j-1]));
时间复杂度分析
时间复杂度很明显为O(
代码
class Solution {public: int minDistance(string word1, string word2) { int m = word1.size(); int n = word2.size(); vector<vector<int> > e(m+1, vector<int>(n+1, 0)); for (int i = 0; i <= m; i++) e[i][0] = i; for (int i = 0; i <= n; i++) e[0][i] = i; int diff = 0; for (int i = 1; i <= m; i++ ) { for (int j = 1; j <= n; j++) { if (word1[i-1] == word2[j-1]) diff = 0; else diff = 1; e[i][j] = min(1 + e[i-1][j], min(1+ e[i][j-1] , diff + e[i-1][j-1])); } } return e[m][n]; }};
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