Partition Array

Question

Problem Statement

Given an array nums of integers and an int k, partition the array (i.e move the elements in "nums") such that:

• All elements < k are moved to the left
• All elements >= k are moved to the right

Return the partitioning index, i.e the first index i nums[i] >= k.

Example

If nums = [3,2,2,1] and k=2, a valid answer is 1.

Note

You should do really partition in array nums instead of just counting the numbers of integers smaller than k.

If all elements in nums are smaller than k, then return nums.length

Challenge

Can you partition the array in-place and in O(n)?

题解1 - 自左向右

C++

class Solution {
public:
int partitionArray(vector<int> &nums, int k) {
int right = 0;
const int size = nums.size();
for (int i = 0; i < size; ++i) {
if (nums[i] < k) {
if (i != right) {
int temp = nums[i];
nums[i] = nums[right];
nums[right] = temp;
}
++right;
}
}

return right;
}
};


题解2 - 两根指针

C++

class Solution {
public:
int partitionArray(vector<int> &nums, int k) {
int left = 0, right = nums.size() - 1;
while (left <= right) {
while (left <= right && nums[left] < k) ++left;
while (left <= right && nums[right] >= k) --right;
if (left <= right) {
int temp = nums[left];
nums[left] = nums[right];
nums[right] = temp;
++left;
--right;
}
}

return left;
}
};