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()];    }