LeetCode 33 — Search in Rotated Sorted Array（C++ Java Python）

題目： http://oj.leetcode.com/problems/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.

題目翻譯：

假設一個有序數組在一個未知的位置旋轉。
（即0 1 2 4 5 6 7可能變成4 5 6 7 0 1 2）。
給定一個目標值進行搜尋。如果數組中存在則返回它的索引，否則返回-1。
可以假設數組中不存在重複的元素。
分析：
        類似二分尋找，但判斷處於哪個子區間要複雜一些。
C++實現：

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

Java實現：

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

Python實現：

class Solution:    # @param A, a list of integers    # @param target, an integer to be searched    # @return an integer    def search(self, A, target):        left = 0        right = len(A) - 1                while left <= right:            mid = (left + right) / 2                        if A[mid] == target:                return mid                        if A[mid] >= A[left]:                if A[mid] > target and A[left] <= target:                    right = mid - 1                else:                    left = mid + 1            else:                if A[mid] < target and A[right] >= target:                    left = mid + 1                else:                    right = mid - 1                    return -1

        感謝閱讀，歡迎評論。