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Given an array nums of integers and an int k, partition the array (i.e move the elements in "nums") such that:
Return the partitioning index, i.e the first index i nums[i] >= k.
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
If nums = [3,2,2,1] and k=2, a valid answer is 1.
Can you partition the array in-place and in O(n)?
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) { //write your code here if(nums == null || nums.length == 0) return 0; int left = 0; int right = nums.length - 1; while(true){ while(left < right && nums[left] < k) left++; while(left < right && nums[right] >= k) right--; if(left == right) break; swap(nums, left, right); } if(nums[left] < k) return left + 1; else return left; } public void swap(int[] nums, int i, int j){ int temp = nums[i]; nums[i] = nums[j]; nums[j] = temp; return; } }
lintcode-medium-Partition Array
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原文地址：http://www.cnblogs.com/goblinengineer/p/5350362.html
