75. Sort Colors

颜色排序算法

最新推荐文章于 2020-07-31 06:04:33 发布

最新推荐文章于 2020-07-31 06:04:33 发布 · 199 阅读

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本文介绍了一种不使用排序函数的颜色排序算法。通过统计不同颜色数量并重新分配数组元素，实现相同颜色相邻且按红、白、蓝顺序排列。此外，还提供了一个一通扫描的优化方案。

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.

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?

我的做法，在本地空间：先统计0，1的数量，确定两个分界点然后对nums赋值就好。

class Solution {
    public void sortColors(int[] nums) {
        int count0=0,count1=0;
        for(int i=0;i<nums.length;i++){
            if(nums[i]==0)
                count0++;
            if(nums[i]==1)
                count1++;
        }
        int index=0;
        for(;index<count0;index++)
            nums[index]=0;
        for(;index<count0+count1;index++)
            nums[index]=1;
        for(;index<nums.length;index++)
            nums[index]=2;
    }
}

参考一种好玩的做法：

如果是0就向左交换，是2就向右移动。

void sortColors(int A[], int n) {
            int second=n-1, zero=0;
            for (int i=0; i<=second; i++) {
                while (A[i]==2 && i<second) swap(A[i], A[second--]);
                while (A[i]==0 && i>zero) swap(A[i], A[zero++]);
            }
        }