Search in rotated Sorted Array

Describe

Suppose a sorted array is rotated on some pivot unknown to you beforehand.

2.1 Array 5

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You is given a target value to search. If found in the array is return its index, otherwise return-1.

Assume no duplicate exists in the array.

Binary.h

#include <iostream> #include <assert.h>class solution{public:int search (int a[], int n, int value) {assert (A)  ; int start = 0;int end = N-1;while (start<=end) {int mid = (End-start)/2 + start;if (a[mid] = = value) return mid;if (A[start]<=a[mid]) {//ordersif (Value<a[mid]&&value>=a[start]) end = Mid-1;elsestart = mid + 1;} Else{//disorderif (Value>a[mid]&&value<=a[end]) start = mid + 1;else{end = Mid-1;}}} return-1;}};

Binary.cpp

#include   "Binary.h" Using namespace std;int main () {int a[9] = { 7,  8, 9, 0, 1, 2, 4, 5, 6 }; Solution s1;cout << s1.search (A, sizeof (a)  / sizeof (A[0]),  7)   << endl;cout << s1.search (A, sizeof (a)  / sizeof (A[0]),  8)   << endl;cout << s1.search (A, sizeof (a)  / sizeof (A[0]),  9)   << endl;cout << s1.search (A, sizeof (a)  / sizeof (A[0]),  0)   << endl;cout << s1.search (A, sizeof (a)  / sizeof (A[0]),  1)   << endl;cout << s1.search (A, sizeof (a)  / sizeof (A[0]),  2)   << endl;cout << s1.search (A, sizeof (a)  / sizeof (A[0]),  4)   << endl;cout << s1.search (a,&Nbsp;sizeof (a)  / sizeof (A[0]),  5)  << endl;cout << s1.search (A,  sizeof (a)  / sizeof (A[0]),  6)  << endl;cout << s1.search (A,  sizeof (a)  / sizeof (A[0]),  3)  << endl;system ("pause"); return 0;}

Operation Result:

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The following is the code for Leetcode_cpp:

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My own program is basically the same as he gave, stating that they still have progress, hehe .... Come on!

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This article is from the "Zero Egg" blog, please be sure to keep this source http://lingdandan.blog.51cto.com/10697032/1771196

Search in rotated Sorted Array