From 3386bf8fc1262cf8898d7957ef9198a0551c0d59 Mon Sep 17 00:00:00 2001
From: ShusenTang <tangshusen@pku.edu.cn>
Date: Wed, 30 Oct 2019 19:41:43 +0800
Subject: [PATCH] Create 274. H-Index.md

---
 solutions/274. H-Index.md | 66 +++++++++++++++++++++++++++++++++++++++
 1 file changed, 66 insertions(+)
 create mode 100644 solutions/274. H-Index.md

diff --git a/solutions/274. H-Index.md b/solutions/274. H-Index.md
new file mode 100644
index 0000000..d33053b
--- /dev/null
+++ b/solutions/274. H-Index.md	
@@ -0,0 +1,66 @@
+# [274. H-Index](https://leetcode.com/problems/h-index/)
+# 思路
+这道题让我们求H指数，如果一个人的学术文章有n篇至少被引用了n次，那么H指数就是满足条件的最大的n。
+
+## 思路一
+根据定义我们可以很好想到先将引用量数组从大到小排个序, 再从后往前遍历直到`i >= citations[i]`成立, 此时i就是H指数. 
+
+## 思路二
+完整的排序复杂度是O(nlogn), 但其实我们没必要排序, 只需要用快排partition的思想不断划分, 例如我们按照从小到大的顺序划分, 
+若某次划分点索引为k:
+* 若 `n - k == citations[k]`, 说明有citations[k]个论文引用量大于等于citations[k], 返回citations[k]即可. 
+* 否则若`n - k > citations[k]`, 说明结果在右边;
+* 否则, 说明结果是citations[k]或左边. 
+
+注意边界条件.
+
+# C++
+## 思路一
+``` C++
+class Solution {
+public:
+    int hIndex(vector<int>& citations) {
+        int n = citations.size();
+        if (!n) return 0;
+        sort(citations.begin(), citations.end(), greater<int>());
+        int i = 0;
+        for (; i < n; i++) 
+            if(i >= citations[i]) break;
+
+        return i;
+    }
+};
+```
+
+## 思路二
+``` C++
+class Solution {
+private:
+    // 标准的快排划分(非常重要)
+    int partition(vector<int> &arr, int left, int right){
+        int pivot = arr[left];
+        while(left < right){
+            while(left < right && pivot <= arr[right]) right--;
+            arr[left] = arr[right]; 
+            while(left < right && pivot >= arr[left]) left++;
+            arr[right] = arr[left];
+        }
+        arr[left] = pivot;
+        return left;
+    }
+public:
+    int hIndex(vector<int>& citations) {
+        int n = citations.size();
+        int left = 0, right = n - 1;
+        while(left <= right){
+            if(left == right) return min(citations[left], n - left);
+            
+            int k = partition(citations, left, right);
+            if(n - k == citations[k]) return citations[k];
+            if(n - k > citations[k]) left = k + 1;
+            else right = k;
+        }
+        return 0;
+    }
+};
+```