Peak Index in a Mountain Array

Peak Index in a Mountain Array

Description

Let’s call an array A a mountain if the following properties hold:

• A.length >= 3
• There exists some 0 < i < A.length - 1 such that A[0] < A[1] < … A[i-1] < A[i] > A[i+1] > … > A[A.length - 1]
  Given an array that is definitely a mountain, return any i such that A[0] < A[1] < … A[i-1] < A[i] > A[i+1] > … > A[A.length - 1].

Example 1:

Input: [0,1,0]
Output: 1

Example 2:

Input: [0,2,1,0]
Output: 1

Tags: Binary Search

解读题意

有这么一个数组，数值会随着下标一直增加，直到顶峰，到顶峰后数值随着下标增加而减少。找出数组中的顶峰（即找数组中的最大值），并返回其下标

思路1

定义指针index，当数组数值不再增加时（到达顶峰）指针会停止，返回指针数值。

class Solution {

   public int peakIndexInMountainArray(int[] A) {

        if (A == null || A.length == 0)
            return 0;

        int index = 0;

        while (A[index] < A[index + 1])
            index++;

        return index;
    }

}

time complexity:O(n)

思路2

利用二叉搜索树来解决问题。该题的数组A[]本身是无序的，但是我们不用返回数组的数值，可以将数组看作A[true,true,true,...,true,false]，判断A[i] < A[i+1]即可。

public class Solution1 {

     public int peakIndexInMountainArray(int[] A) {

        if (A == null || A.length == 0)
            return 0;

        int left = 0, right = A.length - 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (A[mid] < A[mid + 1])
                left = mid + 1;
            else
                right = mid;
        }
        return left;
    }

}

time complexity:O(logn)

leetCode汇总：https://blog.youkuaiyun.com/qingtian_1993/article/details/80588941

项目源码，欢迎star：https://github.com/mcrwayfun/java-leet-code