Find Peak Element
最后更新于：2022-04-02 01:09:03
# Find Peak Element
### Source
- leetcode: [Find Peak Element | LeetCode OJ](https://leetcode.com/problems/find-peak-element/)
- lintcode: [(75) Find Peak Element](http://www.lintcode.com/en/problem/find-peak-element/)
### Problem
A peak element is an element that is greater than its neighbors.
Given an input array where `num[i] ≠ num[i+1]`, find a peak element and returnits index.
The array may contain multiple peaks, in that case return the index to any oneof the peaks is fine.
You may imagine that `num[-1] = num[n] = -∞`.
For example, in array `[1, 2, 3, 1]`, 3 is a peak element and your functionshould return the index number 2.
#### Note:
Your solution should be in logarithmic complexity.
#### Credits:
Special thanks to [@ts](https://oj.leetcode.com/discuss/user/ts) for addingthis problem and creating all test cases.
### 题解1
由时间复杂度的暗示可知应使用二分搜索。首先分析若使用传统的二分搜索，若`A[mid] > A[mid - 1] && A[mid] < A[mid + 1]`，则找到一个peak为A[mid]；若`A[mid - 1] > A[mid]`，则A[mid]左侧必定存在一个peak，可用反证法证明：若左侧不存在peak，则A[mid]左侧元素必满足`A[0] > A[1] > ... > A[mid -1] > A[mid]`，与已知`A[0] < A[1]`矛盾，证毕。同理可得若`A[mid + 1] > A[mid]`，则A[mid]右侧必定存在一个peak。如此迭代即可得解。由于题中假设端点外侧的值均为负无穷大，即`num[-1] < num[0] && num[n-1] > num[n]`, 那么问题来了，这样一来就不能确定峰值一定存在了，因为给定数组为单调序列的话就咩有峰值了，但是实际情况是——题中有负无穷的假设，也就是说在单调序列的情况下，峰值为数组首部或者尾部元素，谁大就是谁了。
备注：如果本题是找 first/last peak，就不能用二分法了。
### Python
~~~
class Solution:
#@param A: An integers list.
#@return: return any of peek positions.
def findPeak(self, A):
if not A:
return -1
l, r = 0, len(A) - 1
while l + 1 < r:
mid = l + (r - l) / 2
if A[mid] < A[mid - 1]:
r = mid
elif A[mid] < A[mid + 1]:
l = mid
else:
return mid
mid = l if A[l] > A[r] else r
return mid
~~~
### C++
~~~
class Solution {
public:
/**
* @param A: An integers array.
* @return: return any of peek positions.
*/
int findPeak(vector A) {
if (A.size() == 0) return -1;
int l = 0, r = A.size() - 1;
while (l + 1 < r) {
int mid = l + (r - l) / 2;
if (A[mid] < A[mid - 1]) {
r = mid;
} else if (A[mid] < A[mid + 1]) {
l = mid;
} else {
return mid;
}
}
int mid = A[l] > A[r] ? l : r;
return mid;
}
};
~~~
### Java
~~~
class Solution {
/**
* @param A: An integers array.
* @return: return any of peek positions.
*/
public int findPeak(int[] A) {
if (A == null || A.length == 0) return -1;
int lb = 0, ub = A.length - 1;
while (lb + 1 < ub) {
int mid = lb + (ub - lb) / 2;
if (A[mid] < A[mid + 1]) {
lb = mid;
} else if (A[mid] < A[mid - 1]){
ub = mid;
} else {
// find a peak
return mid;
}
}
// return a larger number
return A[lb] > A[ub] ? lb : ub;
}
}
~~~
### 源码分析
典型的二分法模板应用，需要注意的是需要考虑单调序列的特殊情况。当然也可使用紧凑一点的实现如改写循环条件为`l < r`，这样就不用考虑单调序列了，见实现2.
### 复杂度分析
二分法，时间复杂度 O(logn)O(\log n)O(logn).
#### Java - compact implementation[leetcode_discussion](#)
~~~
public class Solution {
public int findPeakElement(int[] nums) {
if (nums == null || nums.length == 0) {
return -1;
}
int start = 0, end = nums.length - 1, mid = end / 2;
while (start < end) {
if (nums[mid] < nums[mid + 1]) {
// 1 peak at least in the right side
start = mid + 1;
} else {
// 1 peak at least in the left side
end = mid;
}
mid = start + (end - start) / 2;
}
return start;
}
}
~~~
C++ 的代码可参考 Java 或者 @xuewei4d 的实现。
****> leetcode 和 lintcode 上给的方法名不一样，leetcode 上的为`findPeakElement`而 lintcode 上为`findPeak`，弄混的话会编译错误。
### Reference
- leetcode_discussion
> .
[Java - Binary-Search Solution - Leetcode Discuss](https://leetcode.com/discuss/23840/java-binary-search-solution)[ ↩](# "Jump back to footnote [leetcode_discussion] in the text.")
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