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前端刷力扣 - 189. 旋转数组[中等]

 1 year ago
source link: https://segmentfault.com/a/1190000040678981
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刷题小白遇到这一题:

发现一个有趣,但是不是最好的方法。因为这个方法的效率一般:

但是思路很有意思

如:
nums = [1,2,3,4,5,6,7]
k = 3
可以这么操作。

  1. 把数组整个反转,为: [7,6,5,4,3,2,1]
  2. 从0到k-1反转,为:[5,6,7,4,3,2,1]
  3. 从k到最后反转,为:[5,6,7,1,2,3,4]
/**
 * @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 ? k % nums.length : k;
    
    reverse(nums, 0, nums.length - 1);
    reverse(nums, 0, k - 1);
    reverse(nums, k, nums.length - 1);
};

function reverse(nums, i, j) {
    while(i < j) {
        swap(nums, i, j);
        i++;
        j--;
    }
}

function swap(nums, i, j) {
    let temp = nums[i];
    nums[i] = nums[j];
    nums[j] = temp;
}

题目在这里https://leetcode-cn.com/probl...
有兴趣的同学可以去评论里面看看最高效的实现方式。


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