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?
AC代码
class Solution {
public:
void sortColors(vector<int>& nums) {
int l = 0, r = nums.size() - 1;
for (int i = 0; i <= r; ++i) {
while (nums[i] != 1) {
if (i > r || i < l) break;
else if (nums[i] == 0) swap(nums[i], nums[l++]);
else if (nums[i] == 2) swap(nums[i], nums[r--]);
}
}
}
};
他人代码
class Solution {
public:
void sortColors(vector<int>& nums) {
int red = 0, blue = nums.size() - 1;
for (int i = 0; i <= blue; ++i) {
if (nums[i] == 0) swap(nums[i], nums[red++]);
else if (nums[i] == 2) swap(nums[i--], nums[blue--]);
}
}
};
总结
一次遍历还要O（1）空间复杂度，用双指针，自己写的稍有复杂，别人写的更简单易懂。照抄自：https://www.cnblogs.com/grandyang/p/4341243.html
