33. Search in Rotated Sorted Array

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.

1.我的解答

就是找最大的那个值，然后一分为二，然后target就从这两半中找

class Solution {
public:
int search(int target, vector<int>& num, int begin, int end){
int b = begin, e = end, m = 0;
if((begin > end) || (begin < 0))
return -1;
while(b < e){
m = b + (e - b)/2;
if(target > num[m])
b = m + 1;
else if(target < num[m])
e = m - 1;
else
return m;
}
if(num[b] == target)
return b;
else
return -1;
}

int search(vector<int>& nums, int target) {
int n = nums.size();
int b = 0, e = n - 1, m = 0;
while(e - b > 1){
m = b + (e - b)/2;
if((nums[m] > nums[b]) && (nums[m]) > nums[e])
b = m;
else if((nums[m] < nums[b]) && (nums[m] < nums[e]))
e = m;
else
break;
}
int big = 0;
if(nums[b] > nums[e])
big = b;
else
big = e;
if(target > nums[0])
return search(target,nums,0,big);
else if(target < nums[0])
return search(target,nums,big+1,n-1);
else
return 0;
}
};

2.当然。。。。大神的思路永远比我的好，这里贴上他的解题思路和代码，供以后自己看

永远尊重别人的成果！

Remember the array is sorted, except it might drop at one point.

If nums[0] <= nums[i], then nums[0..i] is sorted (in case of "==" it's just one element, and in case of "<"
there must be a drop elsewhere). So we should keep searching in nums[0..i] if the target lies in this sorted range, i.e., if 

nums[0]
<= target <= nums[i]

.

If nums[i] < nums[0], then nums[0..i] contains a drop, and thus nums[i+1..end] is sorted and lies strictly between nums[i] and nums[0]. So we should keep searching in nums[0..i]
if the target doesn't lie strictly between them, i.e., if 

target
<= nums[i] < nums[0]

 or

nums[i] < nums[0] <= target

Those three cases look cyclic:

nums[0] <= target <= nums[i]
target <= nums[i] < nums[0]
nums[i] < nums[0] <= target

So I have the three checks 

(nums[0] <= target)

, 

(target
<= nums[i])

 and 

(nums[i] < nums[0])

, and I want to know whether exactly
two of them are true. They can't all be true or all be false (check it), so I just need to distinguish between "two true" and "one true". Parity is enough for that, so instead of adding them I xor them, which is a bit shorter and particularly helpful in Java
and Ruby, because those don't let me add booleans but do let me xor them.

代码：

int search(vector<int>& nums, int target) {
int lo = 0, hi = int(nums.size()) - 1;
while (lo < hi) {
int mid = (lo + hi) / 2;
if ((nums[0] > target) ^ (nums[0] > nums[mid]) ^ (target > nums[mid]))
lo = mid + 1;
else
hi = mid;
}
return lo == hi && nums[lo] == target ? lo : -1;
}