題目描述
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.
解題思路
Revised binary search. 鑒于每次要去掉一半，單純對比mid和target無法找到新的起始點，所以需要比較num[start] / num[end] 與mid的關系來確定中位數(shù)在哪半邊。如果在左半邊且start < target < mid，則可以確定新的end = mid - 1；如果在右半邊且mid < target < end，則新的start = mid + 1;
糾結(jié)于到底是start < end 還是 <= 很容易的判斷辦法就是小樣本corner case。如果是 <，那當array只有一個元素且是target的時候，根本進不去loop
Java代碼實現(xiàn)
class Solution {
public int search(int[] nums, int target) {
int start = 0, end = nums.length - 1;
while (start <= end) {
int mid = start + (end - start) / 2;
if (nums[mid] == target) {
return mid;
} else if (nums[mid] > nums[end]) {
// left side sorted
if (target <= nums[mid] && target >= nums[start]) {
end = mid - 1;
} else {
start = mid + 1;
}
} else {
// right side sorted
if (target >= nums[mid] && target <= nums[end]) {
start = mid + 1;
} else {
end = mid - 1;
}
}
}
return - 1;
}
}
