难度
- 简单
- 中等
- 困难
标签
字符串、动态规划题目描述
字符串有三种编辑操作:插入一个字符、删除一个字符或者替换一个字符。 给定两个字符串,编写一个函数判定它们是否只需要一次(或者零次)编辑。示例1:
输入:first = "pale"second = "ple"输出: True
实例2:
输入:first = "pales"second = "pal"输出: False
题解
class Solution {public boolean oneEditAway(String first, String second) {if(first == null || second == null) {return false;}if(first.equals(second)) {return true;}int cha = first.length() - second.length();if(cha < -1 || cha > 1 ) {return false;}if(cha == -1) {return oneEditAway(second, first);}for(int i = 0; i < second.length(); i++) {if(first.charAt(i) != second.charAt(i)) {return first.substring(i + 1).equals(second.substring(cha == 0 ? i + 1 : i));}}return true;}}
