题目：You are given an array nums of non-negative integers. nums is considered special if there exists a number x such that there are exactly x numbers in nums that are greater than or equal to x.

Notice that x does not have to be an element in nums.

Return x if the array is special, otherwise, return -1. It can be proven that if nums is special, the value for x is unique.

 

Example 1:

Input: nums = [3,5]
Output: 2
Explanation: There are 2 values (3 and 5) that are greater than or equal to 2.

Example 2:

Input: nums = [0,0]
Output: -1
Explanation: No numbers fit the criteria for x.
If x = 0, there should be 0 numbers >= x, but there are 2.
If x = 1, there should be 1 number >= x, but there are 0.
If x = 2, there should be 2 numbers >= x, but there are 0.
x cannot be greater since there are only 2 numbers in nums.

Example 3:

Input: nums = [0,4,3,0,4]
Output: 3
Explanation: There are 3 values that are greater than or equal to 3.

Example 4:

Input: nums = [3,6,7,7,0]
Output: -1

方法一：从后面往前找，时间复杂度O(N):

class Solution {
public:
    int specialArray(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        if(nums.size() <= nums[0]) return nums.size();
        for(int i=nums.size()-1; i>0; i--){
            if(nums.size()-i <= nums[i] &&
              nums.size()-i > nums[i-1])
                return nums.size()-i;
        }
        return -1;
    }
};

　　先排序，直接从后面循环找到第一个排位数小于等于当前值且大于前一个值的位置即是答案。

方法二：既然是排序好的，肯定二分法可以解决，捣鼓了好半天，被一个等号给“陷害”了好久~终于完满解决拉：

class Solution {
public:
    int specialArray(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        int start = 0, end = nums.size()-1;
        while(start < end){
            int mid = start + (end - start) / 2;
            if(nums.size()-mid <= nums[mid]) end = mid; // 二分法，重点在这个等号
            else start = mid+1;
        }
        if(start == 0){
            if(nums.size() <= nums[0]) return nums.size();
            else return -1;
        }
        if(nums.size() - start <= nums[start] && nums.size()-start > nums[start-1]) return nums.size()-start;
        return -1;
    }
};