【Leetcode】Search in Rotated Sorted Array

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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.

思路：因為是有序序列的變種，可以考慮應用二分尋找。有兩種分類討論方式，一種是根據A[middle]與target的值比較進行討論；一種是根據找出序列中的遞增序列然後繼續討論。

代碼一：

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

代碼二：

class Solution {public:    int search(int A[], int n, int target) {        if(n <= 0)  return -1;                int left = 0;        int right = n - 1;        int middle = 0;                while(left <= right)        {            middle = (left + right) / 2;                        if(A[middle] == target)                return middle;                        if(A[left] <= A[middle])            {                if(A[left] <= target && target < A[middle])                    right = middle - 1;                else                    left = middle + 1;            }            else            {                if(A[middle] < target && target <= A[right])                    left = middle + 1;                else                    right = middle - 1;            }        }                return -1;    }};