leetcode

leetcode 60题

Posted by franki on June 23, 2021

leetcode 60题 排列序列

给出集合 [1,2,3,…,n]，其所有元素共有 n! 种排列。

按大小顺序列出所有排列情况，并一一标记，当 n = 3 时, 所有排列如下：

  1. “123”
  2. “132”
  3. “213”
  4. “231”
  5. “312”
  6. “321”

给定 n 和 k，返回第 k 个排列。

示例1

输入：n = 3, k = 3
输出："213"

示例2

输入：n = 4, k = 9
输出："2314"

示例3

输入：n = 3, k = 1
输出："123"

思路

每次计算开始和结束的位置

代码

const constArr = [1, 2, 3, 4, 5, 6, 7, 8, 9];
const getNum = function (n, k, arr) {
    let total = constArr.slice(0, n - 1).reduce((total, now) => {
        return total * now;
    }, 1);

    let end = k % total;
    let start = Math.floor(k / total);
    return {
        end: end,
        start: start
    }
}
/**
 * @param {number} n
 * @param {number} k
 * @return {string}
 */
var getPermutation = function (n, k) {
    let newArr = constArr.slice(0, n);
    let str = '';
    let result = getNum(n, k - 1, newArr);
    str += newArr.splice(result.start, 1)[0];
    while (result.end !== 0) {
        result = getNum(--n, result.end, newArr);
        str += newArr.splice(result.start, 1)[0];
    }
    str += newArr.join('');
    return str;
};

复杂度分析:

  • 时间复杂度: O(N)
  • 空间复杂度：O(1)
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