_162.java

package com.fishercoder.solutions;

/**
 * 162. Find Peak Element
 *
 * 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 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 num[-1] = num[n] = -∞.
 For example, in array [1, 2, 3, 1], 3 is a peak element and your function should return the index number 2.*/

public class _162 {

    /**
     * On discuss, this post has very good explanation about an O(logn) solution:
     * https://discuss.leetcode.com/topic/29329/java-solution-and-explanation-using-invariants
     * 
     * Basically, we need to keep this invariant:
     * nums[left] > nums[left-1], then we could return left as the result
     * or nums[right] > nums[right+1], then we could return right as the result
     */
    public static int findPeakElement_Ologn(int[] nums) {

        if (nums == null || nums.length == 0) {
            return 0;
        }
        int left = 0;
        int right = nums.length - 1;
        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] < nums[mid + 1]) {
                left = mid;
            } else {
                right = mid;
            }
        }
        return (left == nums.length - 1 || nums[left] > nums[left + 1]) ? left : right;

    }

    public static void main(String... strings) {
//        int[] nums = new int[]{1,2};
//        int[] nums = new int[]{1};
        int[] nums = new int[]{1, 2, 3, 1};
//        System.out.println(findPeakElement(nums));
        System.out.println(findPeakElement_Ologn(nums));
    }

    /**
     * My original O(n) solution.
     */
    public static int findPeakElement(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int n = nums.length;
        int result = 0;
        for (int i = 0; i < n; i++) {
            if (i == 0 && n > 1 && nums[i] > nums[i + 1]) {
                result = i;
                break;
            } else if (i == n - 1 && i > 0 && nums[i] > nums[i - 1]) {
                result = i;
                break;
            } else if (i > 0 && i < n - 1 && nums[i] > nums[i - 1] && nums[i] > nums[i + 1]) {
                result = i;
                break;
            }
        }
        return result;
    }
}