Given an array nums
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.
We will use the integers 0
, 1
, and 2
to represent the color red, white, and blue, respectively.
You must solve this problem without using the library's sort function.
Example 1:
Input: nums = [2,0,2,1,1,0] Output: [0,0,1,1,2,2]
Example 2:
Input: nums = [2,0,1] Output: [0,1,2]
Example 3:
Input: nums = [0] Output: [0]
Example 4:
Input: nums = [1] Output: [1]
Constraints:
n == nums.length
1 <= n <= 300
nums[i]
is0
,1
, or2
.
Follow up: Could you come up with a one-pass algorithm using only constant extra space?
Companies:
Microsoft, Amazon, Adobe, Apple, Facebook, Bloomberg, Nvidia, Swiggy
Related Topics:
Array, Two Pointers, Sorting
Similar Questions:
// OJ: https://leetcode.com/problems/sort-colors/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
void sortColors(vector<int>& nums) {
vector<int> cnt(3, 0);
for (int n : nums) cnt[n]++;
int i = 0;
for (int j = 0; j < 3; ++j) {
while (cnt[j]--) nums[i++] = j;
}
}
};
// OJ: https://leetcode.com/problems/sort-colors/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
void sortColors(vector<int>& nums) {
int r = 0, g = 0, b = 0;
for (int n : nums) {
if (n == 0) {
nums[b++] = 2;
nums[g++] = 1;
nums[r++] = 0;
} else if (n == 1) {
nums[b++] = 2;
nums[g++] = 1;
} else nums[b++] = 2;
}
}
};
// OJ: https://leetcode.com/problems/sort-colors/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
// Ref: https://leetcode.com/problems/sort-colors/solution/
class Solution {
public:
void sortColors(vector<int>& A) {
int zero = 0, two = A.size() - 1;
for (int i = 0; i <= two; ) {
if (A[i] == 0) {
swap(A[i++], A[zero++]);
} else if (A[i] == 2) {
swap(A[i], A[two--]);
} else ++i;
}
}
};