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.

题解

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。

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).

Java 

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;
    }
}