lintcode: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.

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)?

class Solution {
public:
    int partitionArray(vector<int> &nums, int k) {
        // write your code here
        int left=0;
        int right=nums.size()-1;

        while(left<right){
            while(nums[left]<k) left++;
            while(nums[right]>=k) right--;

            if(left<right){
                int tmp=nums[left];
                nums[left]=nums[right];
                nums[right]=tmp;
                left++;
                right--;
            }
        }

        int res=nums.size();

        for(int i=0;i<nums.size();i++){
            if(nums[i]>=k){
                res=i;
                break;
            }
        }

        return res;

    }
};

代码运行到80%数据出现问题
问题是：

while(nums[left]<k) left++;
while(nums[right]>=k) right--;

处left和right是有可能越界的，导致数组访问出错

AC:

class Solution {
public:
    int partitionArray(vector<int> &nums, int k) {
        // write your code here

        int len=nums.size();

        int left=0;
        int right=len-1;


        while(left<right){
            while(left<len&&nums[left]<k) left++;
            while(right>=0&&nums[right]>=k) right--;

            if(left<right){
                int tmp=nums[left];
                nums[left]=nums[right];
                nums[right]=tmp;
                left++;
                right--;
            }
        }

        int res=len;

        for(int i=0;i<nums.size();i++){
            if(nums[i]>=k){
                res=i;
                break;
            }
        }

        return res;

    }
};