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).

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.

解题:

二分搜索的应用,可以有两种思路:

(1)将数列从mid处划分,按照前半段序列顺序正常,或者不正常,进行条件判断,达到log(n)的二分效果;

(2)将数列从mid处划分,比较mid处元素和target,按照mid元素大,或者mid元素小,进行条件判断,达到log(n)的二分效果;

代码实现1:

代码简介,灵活的运用的二分的思想,将正常/非正常作为二分标准,而不是mid值大还是小。

 class Solution {
public:
int search(vector<int>& nums, int target) {
int start = ;
int end = nums.size() - ;
while (start <= end) {
int mid = (start + end) / ;
if (nums[mid] == target)
return mid;
if (nums[start] <= nums[mid]) {
if (nums[start] <= target && target <= nums[mid])
end = mid - ;
else
start = mid + ;
} else {
if (nums[mid] <= target && target <= nums[end])
start = mid + ;
else
end = mid - ;
}
} return -;
}
};

代码实现2:

思路更常规,虽然代码不如1简介,但是逻辑更好理解。

 class Solution {
public:
int search(vector<int>& nums, int target) {
int left = ;
int right = nums.size() - ;
while (left < right) {
int mid = (left + right) / ;
if (nums[mid] > target) {
if (nums[left] <= target || nums[left] > nums[mid])
right = mid;
else
left = mid + ;
} else if (nums[mid] == target) {
left = mid;
break;
} else {
if (nums[right] >= target || nums[mid] > nums[right])
left = mid + ;
else
right = mid;
}
} if (nums[left] == target)
return left;
else
return -;
}
};

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