[LeetCode 162] Find Peak Element

题目链接：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.
		
		
		Note:
		Your solution should be in logarithmic complexity.
 *
 */

public class FindPeakElement {

	//解法一
//	58 / 58 test cases passed.
//	Status: Accepted
//	Runtime: 214 ms
//	Submitted: 1 minute ago
	
	
	//时间复杂度：O(log(n)) 
	static int peak = -1;
	public int findPeakElement(int[] num) {
		int len = num.length;
		if (len == 1)
			return 0;
		if (num[0] > num[1]) // 最左
			return 0;
		if (num[len - 1] > num[len - 2]) // 最右
			return len - 1;
		findPeakElement(num, 1, len - 2);
		return peak;
	}

	public void findPeakElement(int[] num, int left, int right) {

		if (left <= right) {
			int mid = (right + left) / 2;
			if (num[mid] > num[mid - 1] && num[mid] > num[mid + 1])
				peak = mid;
			else {
				findPeakElement(num, left, mid - 1);
				findPeakElement(num, mid + 1, right);
			}
		}
	}
	
	//解法二
//	58 / 58 test cases passed.
//	Status: Accepted
//	Runtime: 223 ms
//	Submitted: 0 minutes ago
	
	//时间复杂度 O(n)  空间复杂度 O(1)
	 static int findPeakElement1(int[] num) {
		int len = num.length;
		if (len == 1)
			return 0;
		if (num[0] > num[1])				//最左
			return 0;
		if (num[len - 1] > num[len - 2])	//最右
			return len - 1;
		int i = 1;
		for (; i < len - 1; i++) {
			if (num[i] > num[i - 1] && num[i] < num[i + 1])
				break;
		}
		return i;
	 }	

	public static void main(String[] args) {
		System.out.println(findPeakElement1(new int[]{1, 2, 3, 1}));

	}

}

[LeetCode 162] Find Peak Element

原文：http://blog.csdn.net/ever223/article/details/44601405