DP動態規劃專題十三:LeetCode 72. Edit Distance
阿新 • • 發佈:2019-01-07
LeetCode 72. Edit Distance
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')
Example 2: Input: word1 = "intention", word2 = "execution" Output: 5 Explanation: intention -> inention (remove 't') inention -> enention (replace 'i' with 'e') enention -> exention (replace 'n' with 'x') exention -> exection (replace 'n' with 'c') exection -> execution (insert 'u')
每次的操作有三種,也可能沒操作,所以分析case,即可得出結論。
public int minDistance(String word1, String word2) { int n = word1.length(); int m = word2.length(); int[][] dp = new int[n+1][m+1]; for (int i = 1; i <= n; i++) dp[i][0] = i; for (int i = 1; i <= m; i++) dp[0][i] = i; for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { if (word1.charAt(i-1) == word2.charAt(j-1)) { dp[i][j] = dp[i-1][j-1]; } else { dp[i][j] = 1 + Math.min(Math.min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]); } } } for (int i =0; i <= n;i++) { for(int j = 0; j<=m;j++) { System.out.print(dp[i][j] +" "); } System.out.println(); } return dp[n][m]; }
空間優化版:
public int minDistance(String word1, String word2) {
int n = word1.length();
int m = word2.length();
int[] pre = new int[m+1];
int[] cur = new int[m+1];
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= m; j++) {
if (i == 0 && j == 0) cur[j] = 0;
else if (i == 0) cur[j] = j;
else if (j == 0) cur[j] = i;
else{
if (word1.charAt(i-1) == word2.charAt(j-1)) {
cur[j] = pre[j-1];
} else {
cur[j] = 1 + Math.min(Math.min(cur[j-1], pre[j]), pre[j-1]);
}
}
}
int[] tmp = pre;
pre = cur;
cur = tmp;
}
return pre[m];
}