*LeetCode 33. Search in Rotated Sorted Array 二分 值得一练

https://leetcode.com/problems/search-in-rotated-sorted-array/

先说一种相对简单的：首先二分找pivot 然后判断target可能在pivot之前还是之后，然后在可能的范围继续二分

class Solution {
public:
int search(vector<int>& nums, int target) {
if( nums.size() == 0 )return false;
if( nums.size() == 1 ) {
if(target == nums[0])
return 0;
return -1;
}
int pivot;
//unsorted
if( nums[ nums.size()-1 ] < nums[0] ) {
//bisearch
int left = 0, right = nums.size()-1;
while( left <= right ) {
int mid = (left+right) /2 ;
if( nums[mid] >= nums[0] && nums[mid+1] < nums[0]  ){
pivot = mid+1;
break;
} else {
if(nums[mid] < nums[0]) {
right = mid-1;
} else {
left = mid+1;
}
}
}
} else {
return biSearch(nums, 0, nums.size()-1, target);
}
//cout  << "pivot=" << pivot << endl;
int ret ;
if( target > nums[nums.size()-1] ) {
return biSearch(nums, 0, pivot-1, target);
} else {
return biSearch(nums, pivot, nums.size()-1, target );
}
}

int biSearch(vector<int> &nums, int st, int en, int tar) {
int left = st, right = en;
while( left<=right ) {
int mid = (left+right)/2;
if( nums[mid] == tar ) {
//cout << "biSearch st=" << st << "  en=" << en << "  mid=" << mid << endl;
return mid;
}
if( nums[mid] > tar ) {
right = mid - 1;
} else {
left  = mid + 1;
}
}
return -1;
}
};

后来又想想，其实有更好的做法：

假设pivort就是两段数的分界，那么判断mid在哪段，target在哪段，以及mid和target的相对关系，然后二分，代码会更简洁一些：

class Solution {
public:
int search(vector<int>& nums, int target) {
int left=0, right=nums.size()-1, mid=0;
while( left <= right ) {
mid = (left + right) >> 1;
if( nums[mid] == target )return mid;
if( nums[mid] > target ) {
if( nums[mid] >= nums[0] ) {
if( nums[0] > target ) {
left = mid +1;
} else {
right = mid-1;
}
} else {
right = mid-1;
}
} else {
if( nums[mid] >= nums[0] ) {
left = mid+1;
} else {
if( nums[0] > target ) {
left = mid + 1;
} else {
right = mid-1;
}
}
}
}
return -1;
}

};