leetcode-题4

只有一个旋转点，有二分法可以降低复杂度。

二分过后肯定有半边是有序的，我们假设数组是升序的数组，中间有一个旋转点。

解法一：

#include <iostream>
#include <vector>

using namespace std;

int main()
{
    int a[6] = { 11,12,3,4,5,9 };

    int le = 0;
    int r = 5;
    int tar;
    cin >> tar;
    while (le <= r)
    {
        int mid = le + (r - le) / 2;
        if (a[mid] == tar) {
            cout << 1 << ' ';
            break;
        }
        if (a[le] <= a[mid])//the left is order
        {
            if (a[le] <= tar&&tar < a[mid])
                r = mid - 1;
            else
                le = mid + 1;
        }
        else {//the right is order
            if (a[mid] < tar&&tar <= a[r])
                le = mid + 1;
            else
                r = mid - 1;
        }
    }
    if (le > r)cout << -1 << ' ';
    system("pause");
    return 0;
}

解法2：

#include <iostream>
#include <vector>

using namespace std;

int dfs(const vector<int>&num, int le, int re, int tar)
{
	int mid = le + (le + re) / 2;
	if (le > re)return -1;
	if (num[mid] == tar)return 1;
	if (num[le] <= num[mid])
	{
		if (num[le] <= tar&&tar < num[mid])
		{
			re = mid - 1;
			dfs(num, le, re, tar);
		}
		else {
			le = mid + 1;
			dfs(num, le, re, tar);
		}
	}
	else {
		if (num[mid] < tar&&tar < num[re])
		{
			le = mid + 1;
			dfs(num, le, re, tar);
		}
		else
		{
			re = mid - 1;
			dfs(num, le, re, tar);
		}
	}
}

int main()
{
	int tar;
	cin >> tar;
	int a[6] = { 8,9,1,2,3,4 };
	vector<int>num(a,a + 6);
	int res = dfs(num, 0, 5, tar);
	cout << res << endl;
	system("pause");
	return 0;
}