Problem:
Suppose a sorted array 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). How do you find an element in the rotated array efficiently? You may assume no duplicate exists in the array.
Solution:
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public int search(int[] nums, int target) {
int len = nums.length;
int left=0, right=len-1;
// find rotation lenght i.e. index of smallest number in the sort array
while(left < right) {
int mid = (left+right)/2;
if(nums[mid] > nums[right]) {
left = mid+1;
} else {
right = mid;
}
}
// rotation lenght is same as index of left
int rot = left;
// do regular binary search with make use of rotation length
left=0; right=len-1;
while(left<=right) {
int mid = (left+right)/2;
int realMid = (rot+mid)%len;
if(nums[realMid]==target) {
return realMid;
}
else if(nums[realMid]<target) {
left = mid+1;
}
else {
right = mid-1;
}
}
return -1;
}
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