Search in Rotated Sorted Array
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题目要求:
二分查找被旋转的有序数组(不允许数组元素重复),返回待查找数的位置,如果不存在待查找数便返回-1。
例子:
(i.e., 0 1 2 4 5 6 7
被旋转后为 4 5 6 7 0 1 2
),查找0返回4。
代码:
#include<stdafx.h>#include <iostream>using namespace std;class Solution {public:int searchInsert(int A[], int n, int target) {int low=0;int cur=(n-1)/2;int high=n-1;while(low<=high) { int cur= (low+high)/2; if(target == A[cur]) return cur; if(A[cur]<A[high]) { if(target<=A[high] && target>A[cur]) low = cur+1; else high = cur-1; } else { if(target>=A[low] && target<A[cur]) high = cur-1; else low = cur+1; } } return -1; }};void main(){int A[7]={4,5,6,7,0,1,2};Solution s;cout<<s.searchInsert(A,7,0)<<endl;getchar();}
0 0
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