添加内容

website/content/ChapterFour/0274.H-Index.md

@@ -0,0 +1,78 @@
+# [274. H-Index](https://leetcode.com/problems/h-index/)
+
+## 题目
+
+Given an array of citations (each citation is a non-negative integer) of a researcher, write a function to compute the researcher's h-index.
+
+According to the definition of h-index on Wikipedia: "A scientist has index h if h of his/her N papers have at least h citations each, and the other N − h papers have no more than h citations each."
+
+**Example 1**:
+
+```
+
+Input: citations = [3,0,6,1,5]
+Output: 3 
+Explanation: [3,0,6,1,5] means the researcher has 5 papers in total and each of them had 
+             received 3, 0, 6, 1, 5 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.
+
+```
+
+**Note**: 
+
+If there are several possible values for h, the maximum one is taken as the h-index.
+
+
+
+## 题目大意
+
+求 h-index。h-index 值的定义：如果他/她的 N 篇论文中至少有 h 引用，而其他 N-h 论文的引用数不超过 h 引用数。
+
+## 解题思路
+
+可以先将数组里面的数从小到大排序。因为要找最大的 h-index，所以从数组末尾开始往前找，找到第一个数组的值，小于，总长度减去下标的值，这个值就是 h-index。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+// 解法一
+func hIndex(citations []int) int {
+	n := len(citations)
+	buckets := make([]int, n+1)
+	for _, c := range citations {
+		if c >= n {
+			buckets[n]++
+		} else {
+			buckets[c]++
+		}
+	}
+	count := 0
+	for i := n; i >= 0; i-- {
+		count += buckets[i]
+		if count >= i {
+			return i
+		}
+	}
+	return 0
+}
+
+// 解法二
+func hIndex1(citations []int) int {
+	quickSort164(citations, 0, len(citations)-1)
+	hIndex := 0
+	for i := len(citations) - 1; i >= 0; i-- {
+		if citations[i] >= len(citations)-i {
+			hIndex++
+		} else {
+			break
+		}
+	}
+	return hIndex
+}
+
+```