随想录训练营Day56 | DP - - 583. 两个字符串的删除操作, 72. 编辑距离

标签： LeetCode闯关记

583. 两个字符串的删除操作

思路：和动态规划：1143.最长公共子序列 (opens new window)基本相同，只要求出两个字符串的最长公共子序列长度即可

class Solution {
    public int minDistance(String word1, String word2) {
        int len1 = word1.length();
        int len2 = word2.length();
        int[][] dp = new int[len1+1][len2+1];
        //初始化均为0
        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                if(word1.charAt(i-1) == word2.charAt(j-1)){
                    dp[i][j] = dp[i-1][j-1] + 1;
                }else{
                    dp[i][j] = Math.max(dp[i-1][j],dp[i][j-1]);
                }
                //System.out.println("i=" + i + "j=" + j + " dp[i][j]=" + dp[i][j]);
            }
        }
        return len1+len2-2*dp[len1][len2];
    }
}

72. 编辑距离

class Solution {
    public int minDistance(String word1, String word2) {
        //dp[i][j] 表示以下标i-1为结尾的字符串word1，和以下标j-1为结尾的字符串word2，最近编辑距离(最少操作数）为dp[i][j]。
        int len1 = word1.length();
        int len2 = word2.length();
        int[][] dp = new int[len1+1][len2+1];
        for (int i = 0; i <= len1 ; i++) {
            dp[i][0] = i;
        }
        for (int j = 0; j <= len2 ; j++) {
            dp[0][j] = j;
        }
        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2 ; j++) {
                if(word1.charAt(i-1) == word2.charAt(j-1)){
                    dp[i][j] = dp[i-1][j-1];
                }else{
                    dp[i][j] = Math.min(dp[i-1][j] + 1, Math.min(dp[i][j-1] + 1, dp[i-1][j-1]+1));
                }
            }
        }
        return dp[len1][len2];
    }
}