*LeetCode-Find Peak Element

brute force每个和后面的数字比 找到不增长的那个 o(n)

public class Solution {
public int findPeakElement(int[] nums) {
if ( nums == null || nums.length < 2)
return 0;
for ( int i = 0; i < nums.length - 1; i ++ ){
if ( nums[i+1] < nums[i] )
return i;
}
return nums.length-1;
}
}

binary search logn
public class Solution {
public int findPeakElement(int[] nums) {
if ( nums == null || nums.length < 2)
return 0;
int high = nums.length-1;
int low = 0;
while ( low < high ){
int mid = ( low + high ) /2;
int mid2 = mid + 1;
if ( nums[mid2] > nums[mid] ){
low = mid2;
}
else
high = mid;
}
return low;
}
}