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+# [275. H-Index II][title]
+
+## Description
+
+给定一个科研工作者的一系列论文的引用次数（引用次数是非负整数），已经从小到大排好序，请计算他的h因子。
+
+h因子的定义：一个科学家如果有 hh 篇文章的引用次数至少是 hh，并且其他文章的引用次数不超过 hh，我们就说他的影响因子是 hh。
+
+注意： 如果一个人有多个可能的 hh，返回最大的 hh 因子。
+进一步：
+
+这道题目是 LeetCode 274. H-Index的升级版，citations 保证是严格递增的；
+你能否给出时间复杂度是 O(logn)O(logn) 级别的算法？
+
+**Example 1:**
+
+```
+Input: citations = [0,1,3,5,6]
+Output: 3 
+Explanation: [0,1,3,5,6] means the researcher has 5 papers in total and each of them had 
+             received 0, 1, 3, 5, 6 citations respectively. 
+             Since the researcher has 3 papers with at least 3 citations each and the remaining 
+             two with no more than 3 citations each, her h-index is 3.
+```
+
+**Tags:** Math, String
+
+## 题意
+> (二分) O(logn)O(logn)
+  由于数组是从小到大排好序的，所以我们的任务是：
+  在数组中找一个最大的 hh，使得后 hh 个数大于等于 hh。
+  我们发现：如果 hh 满足，则小于 hh 的数都满足；如果 hh 不满足，则大于 hh 的数都不满足。所以具有二分性质。
+  直接二分即可。
+  时间复杂度分析：二分检索，只遍历 lognlogn 个元素，所以时间复杂度是 O(logn)O(logn)
+
+## 题解
+
+### 思路1
+> 。。。。
+
+```go
+package main
+
+func hIndex(citations []int) int {
+	if len(citations) == 0 {
+		return 0
+	}
+	l, r := 0, len(citations)-1
+	for l < r {
+		mid := (l + r) / 2
+		if len(citations)-mid <= citations[mid] {
+			r = mid
+		} else {
+			l = mid + 1
+		}
+	}
+	if citations[l] <= len(citations) {
+		return len(citations) - l
+	}
+	return 0
+}
+```
+
+### 思路2
+> 思路2
+```go
+
+```
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-golang-leetcode][me]
+
+[title]: https://leetcode.com/problems/h-index-ii/
+[me]: https://github.com/kylesliu/awesome-golang-leetcode
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+package Solution
+
+func hIndex(citations []int) int {
+	if len(citations) == 0 {
+		return 0
+	}
+	l, r := 0, len(citations)-1
+	for l < r {
+		mid := (l + r) / 2
+		if len(citations)-mid <= citations[mid] {
+			r = mid
+		} else {
+			
EDBE
l = mid + 1
+		}
+	}
+	if citations[l] <= len(citations) {
+		return len(citations) - l
+	}
+	return 0
+}
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+package Solution
+
+import (
+	"reflect"
+	"strconv"
+	"testing"
+)
+
+func TestSolution(t *testing.T) {
+	//	测试用例
+	cases := []struct {
+		name   string
+		inputs []int
+		expect int
+	}{
+		{"TestCase", []int{0, 1, 3, 5, 6}, 3},
+	}
+
+	//	开始测试
+	for i, c := range cases {
+		t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
+			got := hIndex(c.inputs)
+			if !reflect.DeepEqual(got, c.expect) {
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v",
+					c.expect, got, c.inputs)
+			}
+		})
+	}
+}
+
+//	压力测试
+func BenchmarkSolution(b *testing.B) {
+
+}
+
+//	使用案列
+func ExampleSolution() {
+
+}