LeetCode-Search In Rotated Sorted Array2题解

一.题意描述

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.

数组中可能存在重复数字。

二.解题思路

与简单的二分查找相比，这道题需要多处理一个问题。因为数组是旋转过的排序数组，例如【5 6 6 7 1 2 3 3 4】，那么7为这一数组中的间断点，间断点两侧的数组都是排好序的数组，所以可以直接对这两部分的数组直接使用二分查找。

那么如何找到间断点？我们注意到间断点7大于等于数组中的首个节点5，间断点的下一个节点1小于等于首个节点5，事实上这也正是间断点的特点，我们可以依据这一特点对原数组进行二分查找，直至找到间断点。

在查找间断点的过程中需要注意一个问题，例如数组【1 3 1 1 1】（这也是LeetCode上的一个易错样例），这个数组的中间点1等于第一个点1，他的下一个节点1也等于第一个节点1，然而中间点却并不是间断点。这是一种特殊情况，在这种情况下需要手动将起始节点向后挪动一个，再继续进行二分查找。

在找到间断点之后，判断target数组出现在间断点的左区间还是右区间，问题退化成普通的二分查找，不再赘述。

三.代码实现

class Solution {
public:int findPivot(vector<int>& nums)   　　　　　　//查找中间点的函数{int start = 0;int end = nums.size() - 1;int mid;while(1){int mid = (start + end) / 2;if(start == end)return mid;if(end == start+1){if(nums[end] < nums[start])return start;elsereturn end;}if(nums[mid] == nums[start] && nums[mid + 1] == nums[start])　　　　　　//上文中需要特判的情况{start++;continue;}else if(nums[mid] >= nums[start] && nums[mid + 1] <= nums[start]){return mid;}else if(nums[mid] < nums[start]){end = mid - 1;continue;}else{start = mid + 1;continue;}}}bool binarySearch(vector<int>& nums, int start, int end, int target)　　　　　　　　//普通的二分查找函数{if(end < start)return 0;int mid = (start + end) / 2;if(target == nums[mid])return 1;else if(target < nums[mid])return binarySearch(nums, start, mid - 1, target);elsereturn binarySearch(nums, mid + 1, end, target);}bool search(vector<int>& nums, int target) {int pivot = findPivot(nums);if(!nums.size())return 0;if(target >= nums[0])return binarySearch(nums, 0, pivot, target);elsereturn binarySearch(nums, pivot + 1, nums.size()-1, target);}
};

转载于:https://www.cnblogs.com/tiezhibieek/p/5062069.html

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