一些关于数组查找的题目

//ACM 1 数组
#include <stdio.h>
#include "sort.h"

int arrDump(int * a, int n)
{
	int i = 0;
	printf("Array:[");
	for (i=0; i<n; i++)
	{
		printf(" %d ", a[i]);
	}
	printf("]
");
	return 0;
}

/*
删除有序数组中重复元素 
时间 O(n)
空间 O(1)
*/
int removeSortedDup(int * a, int n, int * len)
{
	int i = 0, index = 0;
	for (i=1; i<n; i++)
	{
		if (a[i] != a[index])//no duplicate
		{
			a[++index] = a[i];//resort
		}
	}
	*len = index+1;
	return 0;
}

/*
允许最多两次重复
时间 O(n)
空间 O(1)
*/
int removeSortedDupTwice(int * a, int n, int * len)
{
	int i = 0, index = 0, sum = 1;
	for (i=1; i<n; i++)
	{
		if (a[i] != a[index])//no duplicate
		{
			a[++index] = a[i];//resort
			sum = 1;
		}
		else if (sum <2)
		{
			a[++index] = a[i];
			sum ++;		
		}
	}
	*len = index+1;
	return 0;
}
/*网上的代码*/
int removeDuplicates(int A[], int n) {
        int index = 2, i=2;
        if (n <= 2) return n;

        for (i = 2; i < n; i++){
            if (A[i] != A[index - 2])//窗口 index-2, index-1
                A[index++] = A[i]; //操作后A[index]始终与之前元素不同
        }

        return index;
    }


/*
二分查找
*/
int BinarySearch(int a[], int len, int target)
{
	int left = 0, right = len-1, mid = 0;
	while (left <= right)
	{
		mid = (left+right) / 2;
		if (target == a[mid])
		{
			return mid;
		}
		else if(target < a[mid])
		{
			right = mid - 1;
		}
		else
		{
			left = mid + 1;
		}
	}
	return -1;
}


/*
有序数组从中间折断，前后两部分换位
Search in Rotated Sorted Array
难度在于二分边界的确定
不使用直接查找，使用二分查找
时间 O(log n)
空间 O(1)
*/
int RotateSearch(int a[], int len, int target)
{
	int left = 0, right = len-1, mid = 0;
	while (left <= right)
	{
		mid = (left+right) / 2;
		if (a[mid] == target) return mid;
		if (a[mid] < a[left])//同a[mid] > a[right]，因为此时 a[left]>a[right]
		{
			if (target>a[mid] && target<=a[right])
				left = mid + 1;
			else
				right = mid - 1;
		}
		else// if (a[mid] >= a[left])
		{
			if (target>=a[left] && target<a[mid])//同target>=a[left] && target<=a[mid]
				right = mid - 1;
			else
				left = mid + 1; 
		}
	}
	return -1;
}

//同上，调用之前的二分查找实现
//int RotateSearch(int a[], int len, int target)
//{
//	int left = 0, right = len-1, mid = 0;
//	while (left < right)
//	{
//		mid = (left+right) / 2;
//		if (a[mid] < a[left])//同a[mid] > a[right]，因为此时 a[left]>a[right]
//		{
//			right = mid;
//		}
//		else if (a[mid] > a[right])
//		{
//			left = mid;
//		}
//		if (left == right-1)
//		{
//			mid = a[left]>a[right] ? left : right;
//			break;//pivot = mid
//		}
//	}
//	printf("Pivot index: %d
", mid);
//	if (a[mid] == target)
//	{
//		return mid;
//	}
//	else if (target > a[0])
//	{
//		return BinarySearch(a, mid+1, target);
//	}
//	else
//	{
//		return (mid+1 + BinarySearch(&a[mid+1], len-mid, target));
//	}
//}

/*
Search in Rotated Sorted Array
允许重复元素，查找数组中是否存在target
注意特殊情况：数组有序但不连续
*/
int RotateSearchDup(int a[], int len, int target)
{
	int left = 0, right = len-1, mid = 0, tmp = -1;
	while (left <= right)
	{
		mid = (left+right) / 2;
		if (a[mid] == target) return mid;
		else if (right-left <= 1)
		{
			return -1;
		}
		if (a[mid] > a[left])//同a[mid] > a[right]，因为此时 a[left]>a[right]
		{
			if (target>=a[left] && target<a[mid])//同target>=a[left] && target<=a[mid]
				right = mid - 1;
			else
				left = mid + 1; 
		}
		else if (a[mid] < a[right])
		{
			if (target>a[mid] && target<=a[right])
				left = mid + 1;
			else
				right = mid - 1;
		}
		else if (a[mid] == a[left])//special situation
		{
			if (target > a[mid])
			{
				tmp = RotateSearchDup(&a[left], mid-left+1, target);
				if (tmp != -1) return left+tmp;
				tmp = RotateSearchDup(&a[mid+1], right-mid, target);
				if (tmp != -1) return mid+1+tmp;
			}
			else left = mid+1;

		}
		else if (a[mid] == a[right])
		{
			if (target > a[mid])
			{
				tmp = RotateSearchDup(&a[mid], right-mid+1, target);
				if (tmp != -1) return mid+tmp;
				tmp = RotateSearchDup(&a[left], mid-left, target);
				if (tmp != -1) return left+tmp;
			}
			else right = mid-1;
		}
	}
	return -1;
}

　　某ACM网站上看到的题目，原题是英文的，今天一回头又找不着了……