Partition Array by Odd and Even
Question
Partition an integers array into odd number first and even number second.
Example
Given [1, 2, 3, 4], return [1, 3, 2, 4]
Challenge
Do it in-place.
Solution
Use two pointers to keep the odd before the even, and swap when necessary.
Java
public class Solution {
/**
* @param nums: an array of integers
* @return: nothing
*/
public void partitionArray(int[] nums) {
if (nums == null) return;
int left = 0, right = nums.length - 1;
while (left < right) {
// odd number
while (left < right && nums[left] % 2 != 0) {
left++;
}
// even number
while (left < right && nums[right] % 2 == 0) {
right--;
}
// swap
if (left < right) {
int temp = nums[left];
nums[left] = nums[right];
nums[right] = temp;
}
}
}
}
C++
void partitionArray(vector<int> &nums) {
if (nums.empty()) return;
int i=0, j=nums.size()-1;
while (i<j) {
while (i<j && nums[i]%2!=0) i++;
while (i<j && nums[j]%2==0) j--;
if (i != j) swap(nums[i], nums[j]);
}
}
Src Code Analysis
Be careful not to forget left < right
in while loop condition.
Complexity
To traverse the array, time complexity is . And maintaining two pointers means space complexity.