https://oj.leetcode.com/problemset/algorithms/
http://siddontang.gitbooks.io/leetcode-solution/content/array/find_peak_element.html
public class Solution {
public int findPeakElement(int[] num) {
// Solution A:
return findPeakElement_Binary(num);
// Solution B:
// return findPeakElement_Gready(num);
}
/////////////////////
// Solution A: Binary
//
public int findPeakElement_Binary(int[] num)
{
int low = 0;
int high = num.length - 1;
while (low <= high)
{
int mid = low + (high - low) / 2;
if ((mid == 0 || num[mid] >= num[mid - 1]) &&
(mid == num.length - 1 || num[mid] >= num[mid + 1]))
{
return mid;
}
else if (mid > 0 && num[mid - 1] >= num[mid])
{
high = mid - 1;
}
else
{
low = mid + 1;
}
}
return num[low];
}
/////////////////////
// Solution A: Gready
//
public int findPeakElement_Gready(int[] num)
{
// Iterate every elements, if it is greater than its neighbours, return the value.
if (num == null || num.length == 0)
return -1; // The input array cannot be null
if (num.length == 1)
return 0; // A edge case.
for (int i = 0 ; i < num.length ; i ++)
{
int cur = num[i];
int pre = i == 0 ? Integer.MIN_VALUE : num[i - 1];
int next = i == num.length - 1 ? Integer.MIN_VALUE : num[i + 1];
if (cur > pre && cur > next)
{
return i;
}
}
// No I did't find it.
// It is possible that all numbers are the same
return -1;
}
}
