描述
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
2.1 数组 5
(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.
You may 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]){//orders
if (value<A[mid]&&value>=A[start])
end = mid - 1;
else
start = mid + 1;
}
else{//disorder
if (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, 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;
}运行结果：
以下是leetcode_cpp的代码：
我自己编的程序基本上和他给的一样，说明自己还是有进步的，嘻嘻。。。。继续加油！
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本文出自 “零蛋蛋” 博客，请务必保留此出处http://lingdandan.blog.51cto.com/10697032/1771196
Search in Rotated Sorted Array
原文：http://lingdandan.blog.51cto.com/10697032/1771196