Java
/**
852. Peak Index in a Mountain Array
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
Note:
3 <= A.length <= 10000
0 <= A[i] <= 10^6
A is a mountain, as defined above.
@tag-array
*/
public class Peak_Index_in_a_Mountain_Array {
class Solution {
public int peakIndexInMountainArray(int[] A) {
int i = 0;
while (A[i] < A[i+1]) {
i++;
}
return i;
}
}
class Solution_BinarySearch {
public int peakIndexInMountainArray(int[] A) {
if (A == null || A.length == 0) {
return 0;
}
int i = 0;
int j = A.length - 1;
while (i < j) {
int mid = (i + j) / 2;
if (mid + 1 < A.length && A[mid + 1] > A[mid]) {
i = mid + 1;
}
else if (mid + 1 < A.length && A[mid + 1] < A[mid]) {
j = mid;
}
else if (mid + 1 < A.length && A[mid - 1] == A[mid]) {
i++;
}
}
return i;
}
}
}Java
class Solution {
public int peakIndexInMountainArray(int[] A) {
int length = A.length;
for (int i = 1; i < length; i++) {
if (A[i] < A[i - 1])
return i - 1;
}
return -1;
}
}