leetcode - 33. Search in Rotated Sorted Array
2017-01-31 10:59
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33. Search in Rotated Sorted Array
Suppose an array sorted inascending order is rotated at some pivot unknown to you beforehand.(i.e.,
0 1 2 4 5 6 7might become
4 5 6 7 0 1 2).
You are given a target value to search. If found in the arrayreturn its index, otherwise return -1.
You may assume no duplicate exists in the array.
在一个旋转数组里查找一个给定的数,并返回查找结果。
若查找成功,就返回该元素的下标,若不成功,就返回-1。
查找问题,一般都是可以用2分法查找,这一题就是毕竟典型的二分查找,只不过它附加上了一些条件。那就是数组是旋转的。旋转的定义,把一个数组最开始的若干个元素搬到数组的末尾,我们称之为数组的旋转。这里是将一组已经排好序的数组进行旋转。
虽然加上了限制的条件,但是仍旧可以利用二分法查找。不妨这样想,假设数组a最大长度为n+1,n>1,不论数组的旋转轴值在哪,a[0]…a[2/n] 和 a[2/n+1]…a
是不是必然有一边是升序的,然后我们只需要在有序的那边调用二分法查找,另外一边就相当于生成了一个子问题,在旋转数组a[2/n+1]…a
里查找一个数字。很清晰地,就可以递归进行求解。
算法思想:
1. 在一个旋转数组里查找一个元素,等于在左右半边数组查找这个元素。
2. 若左半边的是旋转数组跳到第一步,若不是,则调用二分法查找。右半边同左半边一样处理。
PS:判断旋转数组的条件就是A[0] >A
,因为数组原来是升序,A[0] < A
,旋转后大的数就被移动到前面,小的就放在后面。这里的0和N泛指最左边元素和最右边的元素。
int gold = 0;
int result = -1;
void twopart(vector<int>& nums,int left , int right){
if(left > right)return;
int mid = (right + left) / 2;
if(nums[mid] == gold)result = mid; else if(nums[mid] > gold) twopart(nums,left ,mid-1); else twopart(nums,mid+1,right); } void partition(vector<int>& nums,int left , int right){ if(left > right) return; int mid = (right + left) / 2;
if(nums[left] > nums[right]){//判断是否为旋转数组 partition(nums,left,mid); partition(nums,mid+1,right); } else twopart(nums,left,right); } int search(vector<int>& nums, int target) { if(nums.size() < 1)return -1; gold = target; partition(nums,0,nums.size()-1); return result; }
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