leetcode912.排序数组

Posted by serrini on May 30, 2022

Question

给你一个整数数组 nums,请你将该数组升序排列。

示例 1:

输入:nums = [5,2,3,1] 输出:[1,2,3,5]

示例 2:

输入:nums = [5,1,1,2,0,0] 输出:[0,0,1,1,2,5]

提示:

1 <= nums.length <= 5 * 104 -5 * 104 <= nums[i] <= 5 * 104

Related Topics 数组 分治 桶排序 计数排序 基数排序 排序 堆(优先队列) 归并排序 👍 535 👎 0

Answer

void quickSort(vector<int>&nums, int left, int right) {
    if (left >= right) return;
    int privot = nums[left];//privot选择左边的数,让j指针先向左移动
    int i = left;
    int j = right;
    while(i < j) {
        while(nums[j] >= privot && i < j) {
            j--;
        }
        while(nums[i] <= privot && i < j) {
            i++;
        }

        if (i<j) {
            swap(nums[i], nums[j]);
        }
    }
    for(auto it : nums) {
        cout << it << " ";
    }
    cout << endl;

    swap(nums[left], nums[i]);//i==j时

    quickSort(nums, left, i-1);
    quickSort(nums, i+1, right);
}

vector<int> sortArray(vector<int>& nums) {
  	quickSort(nums, 0, nums.size()-1);
 		return nums;
}

//快速排序,从大到小
//AC
void quickSort2(vector<int>&nums, int left, int right) {
    if (left >= right) return;
    int privot = nums[left];
    int i = left;
    int j = right;
    while(i < j) {
        while(nums[j] <= privot && i < j) {
            j--;
        }
        while(nums[i] >= privot && i < j) {
            i++;
        }

        if (i<j) {
            swap(nums[i], nums[j]);
        }
    }

    swap(nums[left], nums[i]);
    quickSort2(nums, left, i-1);
    quickSort2(nums, i+1, right);
}

Attention

  1. privot选择左边的数,让j指针先向左移动,privot选择右边的数,让指针i先向右移动。i==j时交换。