Leetcode-75.Sort Colors
阿新 • • 發佈:2019-04-30
inpu nbsp iter output .so ons swa pass tco
地址:https://leetcode.com/problems/sort-colors/
描述:
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.
實例:
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?
方法1:
// 時間復雜度O(n) // 空間復雜度 O(k) public void sortColors(int[] nums) { int[] count = new int[3]; for (int i = 0; i < nums.length; i++) { if (nums[i] >= 0 && nums[i] <= 2) { count[nums[i]]++; } } int index = 0; for (int i = 0; i < count[0]; i++) { nums[index++] = 0; } for (int i = 0; i < count[1]; i++) { nums[index++] = 1; } for (int i = 0; i < count[2]; i++) { nums[index++] = 2; } }
方法2:
// 時間復雜度O(n) // 空間復雜度 O(1) public void sortColors_2(int[] nums) { int zero = -1; //nums[0,zero]是0 int two = nums.length; // nums[two...n-1]是2 for (int i = 0; i < two; i++) { if (nums[i] == 1) { i++; } else if (nums[i] == 2) { swap(nums, i, --two); } else { if (nums[i] == 0) { swap(nums, ++zero, i++); } } } }
Leetcode-75.Sort Colors