https://leetcode-cn.com/problems/one-away-lcci/
难度中等35收藏分享切换为英文关注反馈
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
示例 2:
输入:
first = "pales"
second = "pal"
输出: False执行用时：128 ms
内存消耗：39.7 MB
/**
* @param {string} first
* @param {string} second
* @return {boolean}
*/
var oneEditAway = function (first, second) {
if (Math.abs(first.length - second.length) > 1) return false;
if (first.length == second.length) { //替换
for (var i = 0; i < first.length; i++) {
if (first[i] != second[i]) {
first = first.slice(0, i) + first.slice(i + 1, first.length);
second = second.slice(0, i) + second.slice(i + 1, second.length);
break;
}
}
}
//first = second + 1
if (first.length - second.length == -1) {
var temp = first;
first = second;
second = temp;
}
if (first.length - second.length == 1) {
for (var i = 0; i < first.length; i++) {
if (first[i] != second[i]) {
first = first.slice(0, i) + first.slice(i + 1, first.length);
}
}
}
if (first == second) return true;
else return false;
};
执行用时：120 ms
内存消耗：39.6 MB
/**
* @param {string} first
* @param {string} second
* @return {boolean}
*/
var oneEditAway = function (first, second) {
if (Math.abs(first.length - second.length) > 1) return false;
for (var i = 0; i < Math.max(first.length,second.length); i++) {
if (first[i] != second[i] && first.length == second.length) {
first = first.slice(0, i) + first.slice(i + 1, first.length);
second = second.slice(0, i) + second.slice(i + 1, second.length);
break;
}
if (first[i] != second[i] && first.length - second.length == 1) {
first = first.slice(0, i) + first.slice(i + 1, first.length);
break;
}
if (first[i] != second[i] && first.length - second.length == -1) {
second = second.slice(0, i) + second.slice(i + 1, second.length);
break;
}
}
if (first == second) return true;
else return false;
};
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