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922.sort-array-by-parity-ii.cpp
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57 lines (53 loc) · 1.34 KB
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// Tag: Array, Two Pointers, Sorting
// Time: O(N)
// Space: O(1)
// Ref: -
// Note: -
// Given an array of integers nums, half of the integers in nums are odd, and the other half are even.
// Sort the array so that whenever nums[i] is odd, i is odd, and whenever nums[i] is even, i is even.
// Return any answer array that satisfies this condition.
//
// Example 1:
//
// Input: nums = [4,2,5,7]
// Output: [4,5,2,7]
// Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
//
// Example 2:
//
// Input: nums = [2,3]
// Output: [2,3]
//
//
// Constraints:
//
// 2 <= nums.length <= 2 * 104
// nums.length is even.
// Half of the integers in nums are even.
// 0 <= nums[i] <= 1000
//
//
// Follow Up: Could you solve it in-place?
//
class Solution {
public:
vector<int> sortArrayByParityII(vector<int>& nums) {
int n = nums.size();
int even = 0;
int odd = 1;
while (even < n && odd < n) {
while (even < n && nums[even] % 2 == 0) {
even += 2;
}
while (odd < n && nums[odd] % 2 == 1) {
odd += 2;
}
if (even < n && odd < n) {
swap(nums[even], nums[odd]);
even += 2;
odd += 2;
}
}
return nums;
}
};