[LeetCode] H-Index II 求H指数之二

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

Hint:

1. Expected runtime complexity is in O(log n) and the input is sorted.

这题是之前那道H-Index 求H指数的拓展，输入数组是有序的，让我们在O(log n)的时间内完成计算，看到这个时间复杂度，应该有很敏锐的意识应该用二分查找法，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，参见代码如下：

class Solution {
public:
    int hIndex(vector<int>& citations) {
        int len = citations.size(), left = 0, right = len - 1;
        while (left <= right) {
            int mid = 0.5 * (left + right);
            if (citations[mid] == len - mid) return mid + 1;
            else if (citations[mid] > len - mid) right = mid - 1;
            else left = mid + 1;
        }
        return len - left;
    }
};

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[LeetCode] H-Index II 求H指数之二

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原文地址：http://www.cnblogs.com/grandyang/p/4782695.html