面试题 01.05. 一次编辑

清疚 2022-11-25 13:07 75阅读 0赞

[面试题 01.05. 一次编辑][01.05.]

[https://leetcode-cn.com/problems/one-away-lcci/][01.05.]

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;
    
    };

[01.05.]: https://leetcode-cn.com/problems/one-away-lcci/