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Array - 75. Sort Colors

Array - 75. Sort Colors

  1. Sort Colors

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?

视频讲解: https://www.bilibili.com/video/BV1mt411J7ER

思路:

在空间复杂度为O(1)的基础上,使用three points做,这三个指针分别代表a类元素和b类元素的分界点,b类元素和未扫描过的元素的分界点,以及位置元素和c类元素的分界点。这样在扫描数组的时候,如果发现当前是a类元素,就把“ab 类指针”和“b未知 指针”上的元素交换,并都向后移动一位。如果是b类元素,就把“b未知指针”向后移动一位,如果当前元素是c类元素,就把“未知c 类指针”上的元素和“b未知 指针”上的元素交换。并把“未知c 类指针”向前移一位。

代码:

java:

class Solution {

    public void sortColors(int[] nums) {
        if (nums == null || nums.length <= 1) return;
      
        int i = 0, j = 0, k= nums.length - 1;
        while (j <= k) {
            if (nums[j] == 0) {
              swap(nums, i++, j++);
            } else if (nums[j] == 1) {
                j++;
            } else {
                swap(nums, j, k--);
            }
        }
    }
  
    private void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}


标题:Array - 75. Sort Colors
作者:michael
地址:https://www.bitbo-liuyang.com/articles/2019/06/25/1561477440163.html

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