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C++实现LeetCode(75.颜色排序)

运维开发网 https://www.qedev.com 2021-07-26 12:41 出处:网络 作者: Grandyang
[LeetCode] 75. Sort Colors 颜色排序 Given an array with n objects colored red, white or blue, sort thwww.cppcns.comem in-place so that objects of the same color are adjacent, with

[LeetCode] 75. Sort Colors 颜色排序

Given an array with n objects colored red, white or blue, sort thwww.cppcns.comem 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:NzDLUOdi0;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:

    www.cppcns.com
  • 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?

这道题的本质还是一道排序的题,题目中给出提示说可以用计数排序,需要遍历数组两遍,那么先来看这种方法,因为数组中只有三个不同的元素,所以实现起来很容易。

- 首先遍历一遍原数组,分别记录 0,1,2 的个数。

- 然后更新原数组,按个数分别赋上 0,1,2。

解法一:

class Solution {
public:
    void sortColors(vector<int>& numNzDLUOdis) {
        vector<int> colors(3);
        for (int num : nums) ++colors[num];
        for (int i = 0, cur = 0; i < 3; ++i) {
            for (int j = 0; j < colors[i]; ++j) {
                nums[cur++] = i;
            }
        }
    }
};

题目中还要让只遍历一次数组来求解,那么就需要用双指针来做,分别从原数组的首尾往中心移动。

- 定义 red 指针指向开头位置,blue 指针指向末尾位置。

- 从头开始遍历原数组,如果遇到0,则交换该值和 red 指针指向的值,并将 red 指针后移一位。若遇到2,则交换该值和 blue 指针指向的值,并将 blue 指针前移一位。若遇到1,则继续遍历。

解法二:

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int red = 0, blue = (int)nums.size() - 1;
        for (int i = 0; i <= blue; ++i) {
            if (nums[i] == 0) {
                swap(nums[i], nums[red++]);
            } else if 编程客栈(nums[i] == 2) {
                swap(nums[i--], nums[blue--]);
            } 
        }
    }
};

当然我们也可以使用 while 循环的方式来写,那么就需要一个变量 cur 来记录当前遍历到的位置,参见代码如下:

解法三:

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int left = 0, right = (int)nums.size() - 1, cur = 0;
        while (cur <= right) {
            if (nums[cur] == 0) {
                swap(nums[cur++], nums[left++]);
            } else if (nums[cur] == 2) {
                swap(nums[cur], nums[right--]);
            } else {
                ++cur;
            }
        }
    }
};

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