LeetCode数组类的题目提交记录 【2】

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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.

eg: 

0 1 2 3 4//右边有序情况
4 0 1 2 3//右边有序情况
3 4 0 1 2//右边有序情况

2 3 4 0 1//左边有序情况
1 2 3 4 0//左边有序情况
**************************************************************/

//代码提交1
//循环法--查找

int search(int* nums, int numsSize, int target) {
    int Left = 0;
    int Right = numsSize - 1;
    int Middle;
    while (Left <= Right)
    {
        Middle = (Left + Right) / 2;

        if (*(nums + Left) == target)
            return Left;
        if (*(nums + Right) == target)
            return Right;
        if (*(nums + Middle) == target)
            return Middle;

        if (*(nums + Left) < *(nums + Middle))
        {
            if (*(nums + Left) <= target && target < *(nums + Middle))
                Right = Middle-1;
            else
                Left = Middle + 1;
            continue;
        }
        else
        {
            if (*(nums + Middle) < target && target <= *(nums + Right))
                Left = Middle + 1;
            else
                Right = Middle-1;
            continue;
        }

    }
    return -1;
}

　　

//代码提交2
//递归法---查找
int searchFun(int* nums, int nLeft,int nRight,int ntargrt) {
    int Left = nLeft;
    int Right = nRight;
    int Middle = (Left + Right) / 2;

    if (Left <= Right)
    {
        if (*(nums + Middle) == ntargrt) {
            return Middle;
        }

        if (*(nums + Middle) < *(nums + Right))
        {
            if (*(nums + Middle) < ntargrt && ntargrt <= *(nums + Right)) 
            {
                Left = Middle + 1;
                return searchFun(nums,Left,Right,ntargrt);
            }
            else            
            {
                Right = Middle - 1;
                return searchFun(nums, Left, Right, ntargrt);
            }
        }

        if (*(nums + Middle) > *(nums + Right)) 
        {
            if (*(nums + Middle) > ntargrt && ntargrt > *(nums + Right))
            {
                Right = Middle - 1;
                return searchFun(nums, Left, Right, ntargrt);
            }
            else 
            {
                Left = Middle + 1;
                return searchFun(nums, Left, Right, ntargrt);
            }
        }

        return -1;
    }

    return -1;
}
int search(int* nums, int numsSize, int target) {
    int Left = 0;
    int Right = numsSize - 1;

    int resultNum = searchFun(nums,Left,Right,target);
    return resultNum;
}

/***********************************************************************
81. Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

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).

Write a function to determine if a given target is in the array.

The array may contain duplicates.

**************************************************************/

bool search(int* nums, int numsSize, int target) {
    int Left = 0;
    int Right = numsSize - 1;
    int Middle;
    while (Left <= Right)
    {
        Middle = (Left + Right) / 2;

        if (*(nums + Left) == target)
            return true;
        if (*(nums + Right) == target)
            return true;
        if (*(nums + Middle) == target)
            return true;

        if (*(nums + Left) < *(nums + Middle))
        {
            if (*(nums + Left) <= target && target < *(nums + Middle))
                Right = Middle - 1;
            else
                Left = Middle + 1;
            continue;
        }
        else if (*(nums + Left) > *(nums + Middle))
        {
            if (*(nums + Middle) < target && target <= *(nums + Right))
                Left = Middle + 1;
            else
                Right = Middle - 1;
            continue;
        }
        else {//*(nums + Left) == *(nums + Middle)则跳过当前位置上的重复
            Left++;
        }

    }
    return false;
}