33.Search in sorted Array

    /*
     * 33.Search in sorted Array 
     * 2016-4-19 by Mingyang 
     * 我自己写的代码，开始没有考虑[3,1]取1得情况，所以现在需要额外的加一个部分来
     * 判断只有2个数的时候
     */
    public static int search1(int[] nums, int target) {
        int len = nums.length;
        if (len == 0 || nums == null)
            return -1;
        return searchHelper(nums, target, 0, len - 1);
    }
    public static int searchHelper(int[] nums, int target, int start, int end) {
        if (start > end)
            return -1;
        int mid = (start + end) / 2;
        //这就是多加的部分
        if(mid==start||mid==end){
            if(nums[start]==target)
                return start;
            if(nums[end]==target)
                return end;
            return -1;
        }
        if (nums[mid] == target) {
            return mid;
        } else if (nums[mid] > nums[start]) {
            if (target >= nums[start] && target <= nums[mid]) {
                return searchHelper(nums, target, start, mid - 1);
            } else {
                return searchHelper(nums, target, mid + 1, end);
            }
        } else {
            if (target >= nums[mid] && target <= nums[end]) {
                return searchHelper(nums, target, mid + 1, end);
            } else {
                return searchHelper(nums, target, start, mid - 1);
            }
        }
    }
    /*
     * 下面是网上的代码，一样的复杂度，可能更具有延展性
     * 如果target比A[mid]值要小------------------------------
     * 如果A[mid]右边有序（A[mid]<A[high]) 那么target肯定不在右边（target比右边的都得小），在左边找
     * 如果A[mid]左边有序,那么比较target和A[low]，如果target比A[low]还要小，
     *证明target不在这一区，去右边找；反之，左边找。
     * 如果target比A[mid]值要大------------------------------- 
     * 如果A[mid]左边有序（A[mid]>A[low]）
     * 那么target肯定不在左边（target比左边的都得大），在右边找 如果A[mid]右边有序
     * 那么比较target和A[high]，如果target比A[high]还要大，证明target不在这一区，去左边找；反之，右边找。
     */
    public int search(int[] A, int target) {
        if (A == null || A.length == 0)
            return -1;
        int low = 0;
        int high = A.length - 1;
        while (low <= high) {    //这里是小于等于哦！！！！！！！！！！！！！----？
            int mid = (low + high) / 2;
            if (target < A[mid]) {
                if (A[mid] < A[high])// right side is sorted
                    high = mid - 1;// target must in left side
                else 
                    if (target < A[low])
// target<A[mid]&&target<A[low]==>means,target cannot be in [low,mid] since this side is sorted
                    low = mid + 1;
                    else
                    high = mid - 1;
            } else if (target > A[mid]) {
                if (A[low] < A[mid])// left side is sorted
                    low = mid + 1;// target must in right side
                else 
                    if (target > A[high])
// right side is sorted. If target>A[high] means target is not in this side
                    high = mid - 1;
                    else
                    low = mid + 1;
            } else
                return mid;
        }        
        return -1;
    }