search-in-rotated-sorted-array
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假设有一个排序的按未知的旋转轴旋转的数组(比如,0 1 2 4 5 6 7 可能成为4 5 6 7 0 1 2)。给定一个目标值进行搜索,如果在数组中找到目标值返回数组中的索引位置,否则返回-1。
你可以假设数组中不存在重复的元素。
样例
给出[4, 5, 1, 2, 3]和target=1,返回 2
给出[4, 5, 1, 2, 3]和target=0,返回 -1
class Solution {public: int search(vector<int> &A, int target) { int n=A.size(); if(n==0) return -1; return helper(A,0,n-1,target); } int helper(vector<int>A,int left,int right,int target){ if(left>right) return -1; if(left==right) return A[left]==target?left:-1; int mid=left+(right-left)/2; if(A[right]==target) return right; if(A[mid]>A[right]&&target>A[right]){ if(A[mid]>target) right=mid-1; else if(A[mid]<target) left=mid+1; else return mid; }else if(A[mid]<A[right]&&target<A[right]){ if(A[mid]>target) right=mid-1; else if(A[mid]<target) left=mid+1; else return mid; }else{ if(A[mid]>target) left=mid+1; else if(A[mid]<target) right=mid-1; else return mid; } return helper(A,left,right,target); }};
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