# 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<int> 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.")