Force deduction solution summary interview question 01.05 Edit once

Directory links ：

Force buckle programming problem - The solution sums up _ Share + Record -CSDN Blog

GitHub Synchronous question brushing items ：

https://github.com/September26/java-algorithms

Original link ： Power button

describe ：

There are three editing operations for Strings : Insert a character 、 Delete or replace a character . Given two strings , Write a function to determine if they only need to be done once ( Or zero times ) edit .

Example  1:

Input : 
first = "pale"
second = "ple"
Output : True
 

Example  2:

Input : 
first = "pales"
second = "pal"
Output : False

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/one-away-lcci
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Their thinking ：

*  Their thinking ：
*  Length of judgement , If second and first The same can only be a replacement . Less than, it can only be deleted , If it is greater than, it can only be plus .
*  Try three possibilities , If none of the three is possible, return false.
*  Then we can simplify the scene , Adding and deleting actually belong to the same scenario , If you reduce one character first Can be converted to second, that second Adding a character can also be converted to first.
*  So we just need to consider reducing first Just a string scene . If second Than first Long , Then the two exchange .

Code ：

public class SolutionMST0105 {

    public boolean oneEditAway(String first, String second) {
        char[] firstChars = first.toCharArray();
        char[] secondChars = second.toCharArray();
        int num = firstChars.length - second.length();
        int diffNum = 0;
        if (num == 0) {
            for (int i = 0; i < firstChars.length; i++) {
                if (firstChars[i] != secondChars[i]) {
                    diffNum++;
                }
            }
            return diffNum <= 1;
        }
        if (Math.abs(num) > 1) {
            return false;
        }
        if (secondChars.length > firstChars.length) {
            char[] local = secondChars;
            secondChars = firstChars;
            firstChars = local;
        }
        for (int i = 0; i < firstChars.length; i++) {
            int secondIndex = i - diffNum;
            if (secondIndex < 0 || secondIndex >= secondChars.length) {
                diffNum++;
            } else if (firstChars[i] != secondChars[secondIndex]) {
                diffNum++;
            }
            if (diffNum > 1) {
                return false;
            }
        }
        return true;
    }
}