LeetCode: Search in Rotated Sorted Array

LeetCode: 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.

地址：https://oj.leetcode.com/problems/search-in-rotated-sorted-array/

算法：将二分查找稍微改进一下就可以了。首先，我们把旋转后的数组分为上部分和下部分，比如题目例子中（4 5 6 7）为上部分，（0 1 2）为下部分，注意上部分和下部分都有可能为空。如果中间节点等于目标值，返回；否则，若中间节点大于尾节点，说明中间节点处于上部分，此时在分三种情况讨论，具体可以看代码；若中间节点小于等于尾节点，说明中间节点处于下部分，此时也可以分三种情况，具体看代码。代码：

class Solution {
public:
    int search(int A[], int n, int target) {
        if(n < 1)   return -1;
        int begin = 0, end = n-1;
        int mid = 0;
        while(begin <= end){
            mid = (begin + end) >> 1;
            if(A[mid] == target){
                return mid;
            }
            if(A[mid] > A[end]){
                if(A[mid] < target)
                    begin = mid + 1;
                else if(target > A[end]){
                    end = mid - 1;
                }else{
                    begin = mid + 1;
                }
            }else{
                if(A[mid] > target){
                    end = mid - 1;
                }else if(target <= A[end]){
                    begin = mid + 1;
                }else{
                    end = mid - 1;
                }
            }
        }
        return -1;
    }
};

第二题：

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

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

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

地址：https://oj.leetcode.com/problems/search-in-rotated-sorted-array-ii/

算法：跟上一题比，多了一个条件：数组中的元素有可能不唯一。同样把bottom和top之间分为上部分和下部分。首先，如果mid就是要找的元素，直接返回；否则，如果top的元素大于bottom的元素，说明bottom和top之间的元素是有序的，则按正常的二分查找即可。否则如果mid大于bottom，则mid位于上部分，所以分三种情况讨论；如果mid小于bottom，则mid为于下部分，所以也分三种情况讨论；剩下mid等于bottom的情况，如果此时mid大于top的话，那说明bottom和mid之间的所有元素都相等，所以都不等与target，故bottom之间设为mid+1；剩下的情况直接线性查找。代码：

class Solution {
public:
    bool search(int A[], int n, int target) {
        if(!A || n < 1){
            return false;
        }
        int bottom = 0;
        int top = n - 1;
        while(bottom <= top){
            int mid = (bottom + top) / 2;
            if(A[mid] == target){
                return true;
            }
            if(A[top] > A[bottom]){
                if(A[mid] < target)
                    bottom = mid + 1;
                else
                    top = mid - 1;
            }else{
                if(A[mid] > A[bottom]){ 
                    if(A[mid] < target)
                        bottom = mid + 1;
                    else if(A[bottom] <= target)
                        top = mid - 1;
                    else
                        bottom = mid + 1;
                }else if(A[mid] < A[bottom]){
                    if(A[mid] > target)
                        top = mid - 1;
                    else if(A[top] >= target)
                        bottom = mid + 1;
                    else
                        top = mid - 1;
                }else{
                    if(A[top] < A[mid]){
                        bottom = mid + 1;
                    }else{
                        for(int i = bottom; i <= top; ++i){
                            if(A[i] == target){
                                return true;
                            }
                        }
                        return false;
                    }
                }
            }
        }
        return false;
    }
};