Question
Formatted question description: https://leetcode.ca/all/1064.html
1064. Fixed Point
Given an array A of distinct integers sorted in ascending order,
return the smallest index i that satisfies A[i] == i. Return -1 if no such i exists.
Example 1:
Input: [-10,-5,0,3,7]
Output: 3
Explanation:
For the given array, A[0] = -10, A[1] = -5, A[2] = 0, A[3] = 3, thus the output is 3.
Example 2:
Input: [0,2,5,8,17]
Output: 0
Explanation:
A[0] = 0, thus the output is 0.
Example 3:
Input: [-10,-5,3,4,7,9]
Output: -1
Explanation:
There is no such i that A[i] = i, thus the output is -1.
Note:
1 <= A.length < 10^4
-10^9 <= A[i] <= 10^9
Algorithm
The basic idea of binary search is to divide n elements into two roughly equal parts, and compare a[n/2] with x.
- If
x=a[n/2], then find x and the algorithm stops; - if
x<a[n/2], as long as you continue to search for x in the left half of array a, - if
x>a[n/2], then as long as you search for x in the right half of array a.
Code
Java
public class Fixed_Point {
// binary search, logN
class Solution {
public int fixedPoint(int[] A) {
//Run binary search
int lo = 0;
int hi = A.length - 1;
while (lo <= hi){
int mi = lo + (hi - lo) / 2;
if (A[mi] == mi) return mi;
if (mi < A[mi]) hi = mi - 1;
else lo = mi+1;
}
return -1;
}
}
class Solution_bruteForce {
public int fixedPoint(int[] A) {
for (int i = 0; i < A.length; i++) {
if (A[i] == i) {
return i;
}
}
return -1;
}
}
}