}     /*      * 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]