From bf043b8b17193061b243b9af1265b1d13dee4e3d Mon Sep 17 00:00:00 2001
From: YDZ <ydz627@gmail.com>
Date: Tue, 1 Oct 2019 14:32:08 +0800
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=20problem=20162?=
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---
 .../162. Find Peak Element.go                 | 43 +++++++++++
 .../162. Find Peak Element_test.go            | 72 +++++++++++++++++++
 Algorithms/0162. Find Peak Element/README.md  | 43 +++++++++++
 3 files changed, 158 insertions(+)
 create mode 100644 Algorithms/0162. Find Peak Element/162. Find Peak Element.go
 create mode 100644 Algorithms/0162. Find Peak Element/162. Find Peak Element_test.go
 create mode 100755 Algorithms/0162. Find Peak Element/README.md

diff --git a/Algorithms/0162. Find Peak Element/162. Find Peak Element.go b/Algorithms/0162. Find Peak Element/162. Find Peak Element.go
new file mode 100644
index 00000000..c36e4a7f
--- /dev/null
+++ b/Algorithms/0162. Find Peak Element/162. Find Peak Element.go	
@@ -0,0 +1,43 @@
+package leetcode
+
+// 解法一 二分
+func findPeakElement(nums []int) int {
+	if len(nums) == 0 || len(nums) == 1 {
+		return 0
+	}
+	low, high := 0, len(nums)-1
+	for low <= high {
+		mid := low + (high-low)>>1
+		if (mid == len(nums)-1 && nums[mid-1] < nums[mid]) || (mid > 0 && nums[mid-1] < nums[mid] && (mid <= len(nums)-2 && nums[mid+1] < nums[mid])) || (mid == 0 && nums[1] < nums[0]) {
+			return mid
+		}
+		if mid > 0 && nums[mid-1] < nums[mid] {
+			low = mid + 1
+		}
+		if mid > 0 && nums[mid-1] > nums[mid] {
+			high = mid - 1
+		}
+		if mid == low {
+			low++
+		}
+		if mid == high {
+			high--
+		}
+	}
+	return -1
+}
+
+// 解法二 二分
+func findPeakElement1(nums []int) int {
+	low, high := 0, len(nums)-1
+	for low < high {
+		mid := low + (high-low)>>1
+		// 如果 mid 较大，则左侧存在峰值，high = m，如果 mid + 1 较大，则右侧存在峰值，low = mid + 1
+		if nums[mid] > nums[mid+1] {
+			high = mid
+		} else {
+			low = mid + 1
+		}
+	}
+	return low
+}
diff --git a/Algorithms/0162. Find Peak Element/162. Find Peak Element_test.go b/Algorithms/0162. Find Peak Element/162. Find Peak Element_test.go
new file mode 100644
index 00000000..f5ef72bb
--- /dev/null
+++ b/Algorithms/0162. Find Peak Element/162. Find Peak Element_test.go	
@@ -0,0 +1,72 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question162 struct {
+	para162
+	ans162
+}
+
+// para 是参数
+// one 代表第一个参数
+type para162 struct {
+	one []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans162 struct {
+	one int
+}
+
+func Test_Problem162(t *testing.T) {
+
+	qs := []question162{
+
+		question162{
+			para162{[]int{2, 1, 2}},
+			ans162{0},
+		},
+
+		question162{
+			para162{[]int{3, 2, 1}},
+			ans162{0},
+		},
+
+		question162{
+			para162{[]int{1, 2}},
+			ans162{1},
+		},
+
+		question162{
+			para162{[]int{2, 1}},
+			ans162{0},
+		},
+
+		question162{
+			para162{[]int{1}},
+			ans162{0},
+		},
+
+		question162{
+			para162{[]int{1, 2, 3, 1}},
+			ans162{2},
+		},
+
+		question162{
+			para162{[]int{1, 2, 1, 3, 5, 6, 4}},
+			ans162{5},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 162------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans162, q.para162
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findPeakElement(p.one))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/0162. Find Peak Element/README.md b/Algorithms/0162. Find Peak Element/README.md
new file mode 100755
index 00000000..9304f0ab
--- /dev/null
+++ b/Algorithms/0162. Find Peak Element/README.md	
@@ -0,0 +1,43 @@
+# [162. Find Peak Element](https://leetcode.com/problems/find-peak-element/)
+
+
+## 题目:
+
+A peak element is an element that is greater than its neighbors.
+
+Given an input array `nums`, where `nums[i] ≠ nums[i+1]`, find a peak element and return its index.
+
+The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.
+
+You may imagine that `nums[-1] = nums[n] = -∞`.
+
+**Example 1:**
+
+    Input: nums = [1,2,3,1]
+    Output: 2
+    Explanation: 3 is a peak element and your function should return the index number 2.
+
+**Example 2:**
+
+    Input: nums = [1,2,1,3,5,6,4]
+    Output: 1 or 5 
+    Explanation: Your function can return either index number 1 where the peak element is 2, 
+                 or index number 5 where the peak element is 6.
+
+**Note:**
+
+Your solution should be in logarithmic complexity.
+
+## 题目大意
+
+峰值元素是指其值大于左右相邻值的元素。给定一个输入数组 nums，其中 nums[i] ≠ nums[i+1]，找到峰值元素并返回其索引。数组可能包含多个峰值，在这种情况下，返回任何一个峰值所在位置即可。你可以假设 nums[-1] = nums[n] = -∞。
+
+说明:
+
+- 你的解法应该是 O(logN) 时间复杂度的。
+
+
+## 解题思路
+
+- 给出一个数组，数组里面存在多个“山峰”，(山峰的定义是，下标 `i` 比 `i-1`、`i+1` 位置上的元素都要大)，找到这个“山峰”，并输出其中一个山峰的下标。
+- 这一题是第 852 题的伪加强版，第 852 题中只存在一个山峰，这一题存在多个山峰。但是实际上搜索的代码是一样的，因为此题只要求随便输出一个山峰的下标即可。思路同第 852 题。