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75.sort-colors.cpp
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74 lines (68 loc) · 1.74 KB
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// Tag: Array, Two Pointers, Sorting
// Time: O(N)
// Space: O(1)
// Ref: -
// Note: -
// Video: https://youtu.be/EA1JjawFqDc
// 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]
//
//
// Constraints:
//
// n == nums.length
// 1 <= n <= 300
// nums[i] is either 0, 1, or 2.
//
//
// Follow up: Could you come up with a one-pass algorithm using only constant extra space?
//
class Solution {
public:
void sortColors(vector<int>& nums) {
int l = 0;
int r = nums.size() - 1;
int i = 0;
while (i <= r) {
if (nums[i] == 0) {
swap(nums[i], nums[l]);
i += 1;
l += 1;
} else if (nums[i] == 2) {
swap(nums[i], nums[r]);
r -= 1;
} else {
i++;
}
}
}
};
class Solution {
public:
void sortColors(vector<int>& nums) {
int l = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] == 0) {
swap(nums[l], nums[i]);
l += 1;
}
}
for (int i = l; i < nums.size(); i++) {
if (nums[i] == 1) {
swap(nums[l], nums[i]);
l += 1;
}
}
}
};