lintcode 31. Partition Array

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https://www.kancloud.cn/kancloud/data-structure-and-algorithm-notes/72953

解法一：** 自左向右**
容易想到的一个办法是自左向右遍历，使用right保存大于等于 k 的索引，i则为当前遍历元素的索引，总是保持i >= right, 那么最后返回的right即为所求。

class Solution {
public:
    /**
     * @param nums: The integer array you should partition
     * @param k: An integer
     * @return: The index after partition
     */
    int partitionArray(vector &nums, int k) {
        // write your code here
        int right = 0;
        for(int i = 0; i < nums.size(); i++){
            if(nums[i] < k && i >= right){
                int tmp = nums[i];
                nums[i] = nums[right];
                nums[right] = tmp;
                right++;
            }
        }
        return right;
    }
}

解法二：两根指针，类似于快速排序

class Solution {
public:
    /**
     * @param nums: The integer array you should partition
     * @param k: An integer
     * @return: The index after partition
     */
    int partitionArray(vector &nums, int k) {
        // write your code here
        int left = 0;
        int right = nums.size() - 1;
        while(left <= right){
            while(nums[right] >= k && right >= left){
                right--;
            }
            while(nums[left] < k && right >= left){
                left++;
            }
            if(left < right){
                int tmp = nums[left];
                nums[left] = nums[right];
                nums[right] = tmp;
            }
        }
        return right+1;
    }
};