75. Find Peak Element 【medium】

75. Find Peak Element 【medium】

There is an integer array which has the following features:

• The numbers in adjacent positions are different.
• A[0] < A[1] && A[A.length - 2] > A[A.length - 1].

We define a position P is a peek if:

A[P] > A[P-1] && A[P] > A[P+1]

Find a peak element in this array. Return the index of the peak.

Notice

• It's guaranteed the array has at least one peak.
• The array may contain multiple peeks, find any of them.
• The array has at least 3 numbers in it.

Example

Given [1, 2, 1, 3, 4, 5, 7, 6]

Return index 1 (which is number 2) or 6 (which is number 7)

Challenge

Time complexity O(logN)

解法一：

 class Solution {
 public:
     /*
      * @param A: An integers array.
      * @return: return any of peek positions.
      */
     int findPeak(vector<int> &A) {
         if (A.empty()) {
             return -;
         }

         int start = ;
         int end = A.size() - ;

         while (start +  < end) {
             int mid = start + (end - start) / ;

             if (A[mid] > A[mid - ]) {
                 if (A[mid] > A[mid + ]) {
                     return mid;
                 }
                 else {
                     start = mid;
                 }
             }
             else {
                 if (A[mid] > A[mid + ]) {
                     end = mid;
                 }
                 else {
                     start = mid;
                 }
             }
         }

         return -;
     }
 };

分类讨论。

解法二：

 class Solution {
     /**
      * @param A: An integers array.
      * @return: return any of peek positions.
      */
     public int findPeak(int[] A) {
         // write your code here
         int start = , end = A.length-; // 1.答案在之间，2.不会出界
         while(start +  <  end) {
             int mid = (start + end) / ;
             if(A[mid] < A[mid - ]) {
                 end = mid;
             } else if(A[mid] < A[mid + ]) {
                 start = mid;
             } else {
                 end = mid;
             }
         }
         if(A[start] < A[end]) {
             return end;
         } else {
             return start;
         }
     }
 }

参考http://www.jiuzhang.com/solution/find-peak-element/的解法，此法更简单。