*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;    }    };