LintCode 31. Partition Array 原创Java参考解答
问题描述
http://www.lintcode.com/en/problem/partition-array/
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.
Notice
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
Example
If nums = [3,2,2,1] and k=2, a valid answer is 1.
解题思路
题意是以k为轴，把比k小的数排到左边，比k大的数排右边，求分界位置。
- 这道题套用quick sort的写法就可以搞定。
参考代码
public class Solution {
/**
*@param nums: The integer array you should partition
*@param k: As description
*return: The index after partition
*/
public int partitionArray(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return 0;
}
int left = 0;
int right = nums.length - 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;
}
}