# C++实现LeetCode(75.颜色排序)

[LeetCode] 75. Sort Colors 颜色排序 Given an array with n objects colored red, white or blue, sort thwww.cppcns.comem in-place so that objects of the same color are adjacent, with

## [LeetCode] 75. Sort Colors 颜色排序

Given an array with n objects colored red, white or blue, sort thwww.cppcns.comem 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:NzDLUOdi0;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]

www.cppcns.com
• 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?

- 首先遍历一遍原数组，分别记录 0，1，2 的个数。

- 然后更新原数组，按个数分别赋上 0，1，2。

```class Solution {
public:
void sortColors(vector<int>& numNzDLUOdis) {
vector<int> colors(3);
for (int num : nums) ++colors[num];
for (int i = 0, cur = 0; i < 3; ++i) {
for (int j = 0; j < colors[i]; ++j) {
nums[cur++] = i;
}
}
}
};```

- 定义 red 指针指向开头位置，blue 指针指向末尾位置。

- 从头开始遍历原数组，如果遇到0，则交换该值和 red 指针指向的值，并将 red 指针后移一位。若遇到2，则交换该值和 blue 指针指向的值，并将 blue 指针前移一位。若遇到1，则继续遍历。

```class Solution {
public:
void sortColors(vector<int>& nums) {
int red = 0, blue = (int)nums.size() - 1;
for (int i = 0; i <= blue; ++i) {
if (nums[i] == 0) {
swap(nums[i], nums[red++]);
} else if 编程客栈(nums[i] == 2) {
swap(nums[i--], nums[blue--]);
}
}
}
};```

```class Solution {
public:
void sortColors(vector<int>& nums) {
int left = 0, right = (int)nums.size() - 1, cur = 0;
while (cur <= right) {
if (nums[cur] == 0) {
swap(nums[cur++], nums[left++]);
} else if (nums[cur] == 2) {
swap(nums[cur], nums[right--]);
} else {
++cur;
}
}
}
};```