[LeetCode]--75. Sort Colors

本文介绍了一种用于将红、白、蓝三种颜色的数组进行排序的算法,通过计数每种颜色的数量,再重新写回数组的方式实现。该算法不使用库函数,适合于颜色种类固定且数量较少的情况。

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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.

click to show follow up.

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?

我第一反应是收集起来,因为只有三个数吗,这样效率就应该蛮高。

public void sortColors(int[] nums) {
        int num0 = 0, num1 = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 0) {
                num0++;
                continue;
            }
            if (nums[i] == 1) {
                num1++;
                continue;
            }
        }

        for (int i = 0; i < nums.length; i++) {
            if (i < num0) {
                nums[i] = 0;
                continue;
            }
            if (i < num0 + num1) {
                nums[i] = 1;
                continue;
            }
            nums[i] = 2;
        }

    }
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