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LeetCode75. Sort Colors(模拟)

2019 年 02 月 02 日 • 阅读: 1273 • LeetCode阅读设置

LeetCode75. Sort Colors(模拟)

  • Medium
  • Accepted:283,496
  • Submissions:691,131

Given an array with n objects colored red, white or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white and blue.

Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.

Note: You are not suppose to use the library's sort function for this problem.

Example:

Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]

Follow up:

  • A rather straight forward solution is a two-pass algorithm using counting sort.
    First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with total number of 0's, then 1's and followed by 2's.
  • Could you come up with a one-pass algorithm using only constant space?

链接

https://leetcode.com/problems/sort-colors/

题意

给定一个含有n个元素的数组,其中只包含0、1、2三种数字,对齐进行排序。不允许使用函数库的排序算法。

题解

因为只包含0,1,2三种数字,所以怎么操作都可以。我的做法是遍历一遍分别统计下0、1、2三种数字的数量然后挨个赋值。时间复杂度$O(n)$,空间复杂度是$O(k)=O(1)​$,k为元素的取值范围。

代码

  • Runtime: 0 ms, faster than 100.00% of C++ online submissions for Sort Colors.
  • Memory Usage: 761.9 KB, less than 68.11% of C++ online submissions for Sort Colors.
class Solution {
public:
    void sortColors(vector<int>& nums) {
        int len = nums.size();
        int cntColor[3] = {0};
        // 先统计
        for (int i = 0; i < len; ++i)
            cntColor[nums[i]]++;
        int cntNum = 0;
        // 后放回
        for (int i = 0; i < 3; ++i)
            for (int j = 0; j < cntColor[i]; ++j)
                nums[cntNum++] = i;
    }
};

有点计数排序的感觉哈。

  • 还有一种做法的思想是三路快速排序,具体见代码。
class Solution {
public:
    void sortColors(vector<int>& nums) {
        int zero = -1; // nums[0, zero] == 0
        int two = nums.size(); // nums[two, n-1] == 2
        for (int i = 0; i < two; )
        {
            if (nums[i] == 1)
                ++i;
            else if (nums[i] == 2)
                swap(nums[i], nums[--two]);
            else if (nums[i] == 0)
                swap(nums[i++], nums[++zero]);
        }
    }
};

The end.

2019年2月2日 星期六