题目描述:

Given an array with n objects colored red, white or blue, sort them 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.

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 an one-pass algorithm using only constant space?

由于数组中元素的值只有三种, 所以可以先遍历一遍数组, 分别记录下三种值的出现次数, 然后再按照值的大小和数量填充到数组中.

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class Solution {
public:
    void sortColors(vector<int>& nums) {
        int colorNum[3] = {0,0,0};
        for(int i = 0; i < nums.size(); i++){
            colorNum[nums[i]]++;
        }
        int k = 0;
        for(int i = 0; i < 3; i++){
            for(int j = 0; j < colorNum[i]; j++){
                nums[k++] = i;
            }
        }
    }
};

关于Follow up中的单次遍历, 可以使用两个指针分别指向red(0)和blue(2)的结尾和开头位置, 然后遍历中间的元素, 通过交换把对应的值放到相应的位置.

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class Solution {
public:
    void sortColors(vector<int>& nums) {
        int redIndex = 0, blueIndex = nums.size() - 1;
        while(nums[redIndex] == 0) redIndex++;
        while(nums[blueIndex] == 2) blueIndex--;
        int i = redIndex;
        while(i <= blueIndex){
            if(nums[i] == 0){
                swap(nums[redIndex++], nums[i]);
                i = max(redIndex, i); // 由于i有可能落后于redIndex, 所以要选择一个较大的值
            }
            else if(nums[i] == 2){
                swap(nums[blueIndex--], nums[i]);
            }
            else{
                i++;
            }
        }
    }
};