Question
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
Solution
Use binary search.
When m and target are on the same side, use binary search rule.
When m and target are on the different side. Select the side where the target belongs.
Code
class Solution {
public int search(int[] nums, int target) {
int l=0;
int r=nums.length-1;
while(l<=r) {
int m = l + (r-l) / 2;
if(nums[m] == target) return m;
if(target < nums[l]) {
// target is on the right side
if(nums[m] >= nums[l]) {
// m is on the left side
l = m+1;
}
else {
// m is also on the right side
if(nums[m] > target) {
r = m - 1;
}
else {
l = m + 1;
}
}
}
else {
// target is on the left side
if(nums[m] >= nums[l]) {
// m is also on the left side
if(nums[m] > target) {
r = m - 1;
}
else {
l = m + 1;
}
}
else {
r = m - 1;
}
}
}
return -1;
}
}
Performance
O(n) time complexity.