275. H-Index II

            }
     /*
      * 275. H-Index II
      * 1.3 by Mingyang
      * 在Sort里面有线性解法的
      * 输入数组是有序的，让我们在O(log n)的时间内完成计算，
      * 我们最先初始化left和right为0和数组长度len-1，然后取中间值mid，
      * 比较citations[mid]和len-mid做比较，如果前者大，则right移到mid之前，
      * 反之right移到mid之后，终止条件是left>right，最后返回len-left即可
      * for paper[m]. there should be at least (len – m) papers with citations >= citations[m]
      */
     public static int hIndex(int[] citations) {
            if (citations == null || citations.length == 0) {
                return 0;
            }

            int low = 0;
            int high = citations.length - 1;
            int len = citations.length;
            while (low <= high) {
                int mid = (low + high)/2;
                if (citations[mid] == len - mid) return len - mid;      //第k个数等于k，绝配----len-mid表示这个数从大到小排在第几位
                else if (citations[mid] > len - mid) high = mid - 1;
                else low = mid + 1;
            }
            return len - low;
        }
     //孟严的方法
     public int hIndex2(int[] citations) {
            int len=citations.length;
            if(len==0)
             return 0;
            int left=0,right=len-1;
            while(left<=right)
            {
                int mid=left+(right-left)/2;
               // 0 4 5 6 7   5>=3
                 //4>=4
                if(citations[mid]>=len-mid) //--------mid=2
                {
                    if(mid==0||citations[mid-1]<len-mid+1)
                        return len-mid;
                    else
                        right=mid-1;  // 0 4
                        //
                }
                else
                 left=mid+1;
            }
            return 0;
     }