[leetcode]Search in Rotated Sorted Array

有人用了很复杂的讨论-，-

我感觉我是考虑不了那么多。。

那么我先二分找到那个分割点，然后。。。分两段来二分找

class Solution {
public:
    int bs(int A[] , int start , int end , int target){
        int l = start;
        int r = end;
        while(l <= r){
            int mid = (l + r) / 2;
            if(A[mid] == target) return mid;
            if(A[mid] < target) l = mid + 1;
            else r = mid - 1;
        }
        return -1;
    }
    int search(int A[], int n, int target) {
        if(n <= 0) return -1;
        int l = 0;
        int r = n - 1;
        while(l < r)
        {
            int mid = (l + r) / 2;
            if(A[mid] > A[l]) l = mid;
            else r = mid;
        }
        //split l
        int idx = bs(A , 0 , l , target);
        if(idx != -1) return idx;
        return bs(A , l + 1 , n-1, target);
    }
};

 ------update 07/20/2014----

 我也讨论下，其实就几种情况

mid，target在左边还是右边的区域，然后该怎么做。

class Solution {
public:
    int search(int A[], int n, int target) {
        int l = 0;
        int r = n - 1;
        while(l <= r) {
            int mid = (r-l)/2+l;
            if (A[mid] == target) return mid;
            if (A[mid] >= A[l]) {
                //left
                if (target < A[l]) {
                    //target is on the right
                    l = mid + 1;
                } else {
                    //target is on the left
                    if (A[mid] < target) {
                        l = mid + 1;
                    } else {
                        r = mid - 1;
                    }
                }
            } else {
                //right
                if (target <= A[r]) {
                    //target is on the right
                    if (A[mid] < target) {
                        l = mid + 1;
                    } else {
                        r = mid - 1;
                    }
                } else {
                    r = mid - 1;
                }
            }
        }
        return -1;
    }
};