Follow up for H-Index: What if the citations
array is sorted in ascending order? Could you optimize your algorithm?

一刷
令len-h = left, 则N-left = h
用binary search在数组中寻找，nums[left]为小于h和大于h的交界。return len-left

public class Solution {
    public int hIndex(int[] citations) {
       int left = 0, len = citations.length, right = len-1, mid;
        while(left<=right){
            mid = left + (right-left)/2;
            if(citations[mid]>=(len-mid)) right = mid-1;
            else left = mid+1;
        }
        return len-left;
    }
}

二刷
思路和1刷相同

public class Solution {
    public int hIndex(int[] citations) {
        int left = 0, len = citations.length, right = len-1, mid;
        while(left<=right){
            mid = left + (right-left)/2;
            if(citations[mid] >(len-mid)) right = mid-1;
            else if(citations[mid] <(len-mid)) left = mid+1;
            else return len-mid;
        }
        return len - left;
    }
}

三刷
同上

public class Solution {
    public int hIndex(int[] citations) {
        int left = 0, len = citations.length, right = len-1, mid;
        while(left<=right){
            mid = left + (right-left)/2;
            if(citations[mid] >=(len-mid)) right = mid-1;
            else if(citations[mid] <(len-mid)) left = mid+1;
        }
        return len - left;
    }
}