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162. Find Peak Element

时间:2015-04-14 12:39:23      阅读:218      评论:0      收藏:0      [点我收藏+]

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.

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Note:

Your solution should be in logarithmic complexity.

二分查找,如果当前只大于两侧,当前值即为所找。否则如果左侧num[mid-1] > num[mid],则应在左侧,因为如果num[mid-2] < num[mid-1]则num[mid-1]即为所找,否则继续向左找,直到num[-1]负无穷。

public class Solution {
  public int findPeakElement(int[] num) {
    int left = 0;
    int right = num.length - 1;
    while (left <= right) {
      int mid = (left + right) / 2;
      if ((mid == 0 || num[mid] > num[mid - 1]) &&
        (mid == num.length - 1 || num[mid] > num[mid + 1])) {
        return mid;
      }
      if (mid > 0 && num[mid] < num[mid - 1]) {
        right = mid -1;
      } else {
        left = mid + 1;
      }
    }
    return -1;
  }
}

162. Find Peak Element

原文:http://www.cnblogs.com/shini/p/4424367.html

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