# LeetCode75. Sort Colors（模拟）

## LeetCode75. Sort Colors（模拟）

• Medium
• Accepted：283,496
• Submissions：691,131

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]

• 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?

### 链接

https://leetcode.com/problems/sort-colors/

### 代码

• Runtime: 0 ms, faster than 100.00% of C++ online submissions for Sort Colors.
• Memory Usage: 761.9 KB, less than 68.11% of C++ online submissions for Sort Colors.
class Solution {
public:
void sortColors(vector<int>& nums) {
int len = nums.size();
int cntColor[3] = {0};
// 先统计
for (int i = 0; i < len; ++i)
cntColor[nums[i]]++;
int cntNum = 0;
// 后放回
for (int i = 0; i < 3; ++i)
for (int j = 0; j < cntColor[i]; ++j)
nums[cntNum++] = i;
}
};

• 还有一种做法的思想是三路快速排序，具体见代码。
class Solution {
public:
void sortColors(vector<int>& nums) {
int zero = -1; // nums[0, zero] == 0
int two = nums.size(); // nums[two, n-1] == 2
for (int i = 0; i < two; )
{
if (nums[i] == 1)
++i;
else if (nums[i] == 2)
swap(nums[i], nums[--two]);
else if (nums[i] == 0)
swap(nums[i++], nums[++zero]);
}
}
};

The end.

2019年2月2日 星期六