算法第九周作业01

Description

Edit Distance

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

Solution

动态规划dp[word1.length()+1][word2.length()+1]，其中dp[i][j]表示word1的第i个元素和word2的第j个元素对应时需要调整的次数（例如word1=”abef”，word2=”abcd”中，dp[4][4]表示word1转换成word2需要调整的次数，而dp[3][3]表示”abe”转换成”abc”需要调整的次数）

当word1[i] == word2[j]时，dp[i][j] = dp[i-1][j-1]

当word1[i] ! = word2[j]时，dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1，分别对应增、删、改

初始化，dp[i][0] = i; //对应word1.subStr(0, i)转换成”“（空字符串）时需要调整（删除）的次数

dp[0][j] = j; //对应”“转换成word1.subStr(0, i)时需要调整（插入）的次数

Code

public int minDistance(String word1, String word2) {
// 构建dp数组，注意大小
int[][] dp = new int[word1.length() + 1][word2.length() + 1];
// 纵向初始化
for (int i = 0; i <= word1.length(); i++) {
dp[i][0] = i;
}
// 横向初始化
for (int i = 0; i <= word2.length(); i++) {
dp[0][i] = i;
}
int tmp;
// 遍历求解dp元素
for (int i = 1; i <= word1.length(); i++) {
for (int j = 1; j <= word2.length(); j++) {
if(word1.charAt(i-1) == word2.charAt(j-1)){
// 对应的字符相等时
dp[i][j] = dp[i-1][j-1];
} else {
// 对应的字符不相等时
tmp = Math.min(dp[i][j-1], dp[i-1][j]);
dp[i][j] = Math.min(dp[i-1][j-1], tmp) + 1;
}
}
}
// 最后一下即为所求
return dp[word1.length()][word2.length()];
}