leetcode--33. Search in Rotated Sorted Array

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.

我的方法：
与discussion中第二种一致。
step1：确定pivot在哪一侧
step2：根据pivot所在边的特点，进行mid的确定
这里格外注意边界的问题：
if(nums[mid] < nums[tail]) { //pivot is on the left

if(tar > nums[tail] || tar < nums[mid]) tail = mid;

都是需要注意的点。否则会引起死循环。

class Solution {
public:
int search(vector<int>& nums, int tar) {

int len = nums.size();
if(len == 0) return -1;
int head = 0;
int tail = len-1;

while(head < tail){
int mid = head + (tail-head)/2;
if(nums[mid] == tar) return mid;
if(nums[mid] < nums[tail]) { //pivot is on the left
if(tar > nums[tail] || tar < nums[mid]) tail = mid;
else head = mid+1;
}
else{
if(tar < nums[head] || tar > nums[mid]) head = mid+1;
else tail = mid;
}
}
return nums[tail] == tar? tail : -1;
}
};

方案二：
discuss中的第一种方法：
step1：用一个while循环，找到pivot的坐标
step2：还原原始下标，二分法查找
int realmid=(mid+rot)%n;
class Solution {
public:
int search(int A[], int n, int target) {
int lo=0,hi=n-1;
// find the index of the smallest value using binary search.
// Loop will terminate since mid < hi, and lo or hi will shrink by at least 1.
// Proof by contradiction that mid < hi: if mid==hi, then lo==hi and loop would have been terminated.
while(lo<hi){
int mid=(lo+hi)/2;
if(A[mid]>A[hi]) lo=mid+1;
else hi=mid;
}
// lo==hi is the index of the smallest value and also the number of places rotated.
int rot=lo;
lo=0;hi=n-1;
// The usual binary search and accounting for rotation.
while(lo<=hi){
int mid=(lo+hi)/2;
int realmid=(mid+rot)%n;
if(A[realmid]==target)return realmid;
if(A[realmid]<target)lo=mid+1;
else hi=mid-1;
}
return -1;
}
};