Find Peak Element详解编程语言

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.

Your solution should be in logarithmic complexity.

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。

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; 

};

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; 

}

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.

二分法，时间复杂度 O(logn).

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; 

}

原创文章，作者：ItWorker，如若转载，请注明出处：https://blog.ytso.com/20656.html