[LintCode] 31 Partition Array 解题报告

Description
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.



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



思路
标准的3-way-partition。



Code

class Solution {
    /**
     * @param nums: A list of integer which is 0, 1 or 2 
     * @return: nothing
     */
    public void sortColors(int[] nums) {
        // write your code here
        
        if (nums == null || nums.length < 1) {
            return;
        }
        
        int lt = 0;
        int gt = nums.length - 1;
        int v = 1;
        int i = 0;
        while (i <= gt) {
            if (nums[i] < v) {
                exch(nums, lt++, i++);
            }
            else if (nums[i] > v) {
                exch(nums, i, gt--);
            }
            else {
                i++;
            }
        }
    }
    
    public void exch(int[] nums, int x, int y) {
        int tmp = nums[x];
        nums[x] = nums[y];
        nums[y] = tmp;
    }
}