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旋转数组 #7

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Aras-ax opened this issue Mar 4, 2019 · 0 comments
Open

旋转数组 #7

Aras-ax opened this issue Mar 4, 2019 · 0 comments

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@Aras-ax
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Aras-ax commented Mar 4, 2019

给定一个数组,将数组中的元素向右移动k个位置,其中k是非负数。

示例 1:

输入: [1,2,3,4,5,6,7] 和 k = 3
输出: [5,6,7,1,2,3,4]
解释:
向右旋转 1 步: [7,1,2,3,4,5,6]
向右旋转 2 步: [6,7,1,2,3,4,5]
向右旋转 3 步: [5,6,7,1,2,3,4]

示例 2:

输入: [-1,-100,3,99] 和 k = 2
输出: [3,99,-1,-100]
解释: 
向右旋转 1 步: [99,-1,-100,3]
向右旋转 2 步: [3,99,-1,-100]

说明:

  1. 尽可能想出更多的解决方案,至少有三种不同的方法可以解决这个问题。
  2. 要求使用空间复杂度为 O(1) 的原地算法。

解答

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var rotate = function(nums, k) {
    k = k % nums.length;
    reverse(nums, 0, nums.length - 1);
    reverse(nums, 0, k - 1);
    reverse(nums, k, nums.length - 1);
};

function reverse(nums, start, end){
    while(start < end){
        nums[start] = nums[start] ^ nums[end];
        nums[end] = nums[start] ^ nums[end];
        nums[start] = nums[start] ^ nums[end];
        start++;
        end--;
    }
}
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