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]
Solution_01:
#include <iostream>
#include <vector>
#include <cassert>
using namespace std;
class Solution{
public:
// 时间复杂度: o(n)
// 空间复杂度: o(1)
void sortColors(vector<int>& nums){
int count[3] = {0}; // 存放0,1,2三个元素的频率
for(int i = 0; i < nums.size(); i++){
assert( nums[i] >=0 && nums[i] <=2 ); //防止数组越界
count[nums[i]] ++; // 一定要学会这种用法,用于计数
}
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;
}
}
};
int main(){
return 0;
}
Solution_02
#include <iostream>
#include <vector>
#include <cassert>
using namespace std;
class Solution{
public:
// 时间复杂度: o(n)
// 空间复杂度: o(1)
// 只遍历了一遍数组
// 用的是三路快排的思想
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) {
two --;
swap(nums[i], nums[two]);
} else{ // nums[i] == 0
zero ++;
swap(nums[zero], nums[i]);
i++;
}
}
}
};
int main(){
return 0;
}