【LeetCode】33. Search in Rotated Sorted Array (4 solutions)

Search in Rotated Sorted Array

Suppose a sorted array 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.

 

与Search in Rotated Sorted Array II,Find Minimum in Rotated Sorted Array,Find Minimum in Rotated Sorted Array II对照看

解法一：顺序查找 just a joke :D

class Solution {
public:
    int search(int A[], int n, int target) {
        for(int i = 0; i < n; i ++)
        {
            if(A[i] == target)
                return i;
        }
        return -1;
    }
};

 

解法二：二分查找

先用二分法找到最大元素，将数组切分成两个有序数组，再进行二分查找

class Solution {
public:
    int search(int A[], int n, int target) {
        if(n==1)
            return (target==A[0])?0:-1;
            
        //find the maximum first
        int low = 0;
        int high = n-1;
        while(low < high)
        {
            int mid = (low+high)/2;
            if(A[mid] < A[low])
                high = mid-1;
            else if(A[mid] > A[low])
                low = mid;
            else
            {//low+1==high
                if(A[high]>A[low])
                    low = high;
                break;
            }
        }
        int ind = low;
        //to here, low is the index of maximum
        //0~ind, ind+1~n-1 are two sorted arrays
        if(target >= A[0])
        {//first array: 0~ind
            low = 0;
            high = ind;
            while(low <= high)
            {
                int mid = (low+high)/2;
                if(target == A[mid])
                    return mid;
                else if(target > A[mid])
                    low = mid+1;
                else
                    high = mid-1;
            }
            return -1;
        }
        else
        {//second array: ind+1, n-1
            low = ind+1;
            high = n-1;
            while(low <= high)
            {
                int mid = (low+high)/2;
                if(target == A[mid])
                    return mid;
                else if(target > A[mid])
                    low = mid+1;
                else
                    high = mid-1;
            }
            return -1;
        }
    }
};

 

解法三：可处理重复元素的二分查找，即不断去掉low与high元素

class Solution {

public:
    int search(int A[], int n, int target) {
        int low = 0;
        int high = n-1;
        while (low <= high)
        {
            int mid = (low+high)/2;
            if(A[mid] == target) 
                return mid;
            if (A[low] < A[mid])
            {
                if(A[low] <= target && target < A[mid])
                //binary search in sorted A[low~mid-1]
                    high = mid - 1;
                else
                //subproblem from low to high
                    low = mid + 1;
            }
            else if(A[mid] < A[high])
            {
                if(A[mid] < target && target <= A[high])
                //binary search in sorted A[mid+1~high]
                    low = mid + 1;
                else
                //subproblem from low to mid-1
                    high = mid - 1;
            }
            else if(A[low] == A[mid])
                low += 1;    //A[low]==A[mid] is not the target, so remove it
            else if(A[mid] == A[high])
                high -= 1;  //A[high]==A[mid] is not the target, so remove it
        }
        return -1;
    }
};

 

解法四：

二分查找，先对mid元素处于前半段还是后半段分情况讨论，再对target元素处于前半段还是后半段分情况讨论。

class Solution {
public:
    int search(int A[], int n, int target) {
        return search(A, 0, n-1, target);
    }
    int search(int A[], int left, int right, int target)
    {
        if(left > right)
            return -1;
        
        if(A[left] < A[right])
        // one part, binary search
            return binarySearch(A, left, right, target);
            
        // else, two part
        int mid = left + (right-left) / 2;  //prevent overflow
        if(A[mid] == target)
            return mid;
        else if(A[left] > A[mid])
        {// mid is in second part
            if(target > A[mid])
            {// target may be in the first part (case1), or second part after mid(case2)
                if(target == A[left])
                    return left;
                else if(target > A[left])
                {// case1
                    return search(A, left, mid-1, target);
                }
                else
                {// case2
                    return search(A, mid+1, right, target);
                }
            }
            else
            {// target is in the second part before mid
                return search(A, left, mid-1, target);
            }
        }
        else
        {// mid is in first part
            if(target > A[mid])
            {// target is in first part after mid
                return search(A, mid+1, right, target);
            }
            else
            {// target may be in the first part before mid (case1), or second part
                if(target == A[left])
                    return left;
                else if(target > A[left])
                {// case1
                    return search(A, left, mid-1, target);
                }
                else
                {// case2
                    return search(A, mid+1, right, target);
                }
            }
        }
    }
    int binarySearch(int A[], int left, int right, int target)
    {
        while(left <= right)
        {
            int mid = left + (right-left) / 2;  //prevent overflow
            if(A[mid] == target)
                return mid;
            else if(A[mid] > target)
                right = mid - 1;
            else
                left = mid + 1;
        }
        return -1;
    }
};