LeetCode 1.Two Sum

最新推荐文章于 2020-10-21 10:33:51 发布

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本文介绍了一种解决两数之和问题的C语言实现方法。通过先排序再查找的策略，在给定的整数数组中寻找两个数，使它们的和等于特定的目标值。文章提供了详细的代码实现，采用归并排序算法进行排序，最终返回这两个数的下标。

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Two Sum

Given an array of integers, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution.
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

查看了好多人的博客，都是用C++写的，几乎也都是用的hash表。俺们C++一直都不自信，所以都用C写代码。

本题的思路是先排序，然后再查找。由于用的是归并排序算法，所以时间复杂度是O(NlogN)。下面是C代码：

void Merge(int *A, int p, int q, int r)
{
	int *arr = (int *)malloc(sizeof(int)*(r - p + 1));
	memset(arr, 0, sizeof(int)*(r - p + 1));
	int i = p,j=q+1,k=0;
	while (i <= q&&j <= r)
	{
		arr[k++] = (A[i] <= A[j]) ? A[i++] : A[j++];
	}
	while (i <= q)
	{
		arr[k++] = A[i++];
	}
	while (j <= r)
	{
		arr[k++] = A[j++];
	}

	for (i = 0; i < k; i++)
	{
		A[p + i] = arr[i];
	}
	free(arr);
}


void MergeSort(int *A, int p, int r)
{
	int q = (p + r) / 2;
	if (p < r)
	{
		MergeSort(A, p, q);
		MergeSort(A, q + 1, r);
		Merge(A, p, q, r);
	}
}

int* twoSum(int* nums, int numsSize, int target)
{
	if (nums == NULL)
	{
		return NULL;
	}

	int *tmp = (int *)malloc(sizeof(int)*numsSize);
	memset(tmp, 0, sizeof(int)*numsSize);

	int k = 0;
	while (k<numsSize)
	{
		tmp[k] = nums[k];
		k++;
	}

	MergeSort(tmp, 0, numsSize - 1);

	int *index = (int *)malloc(sizeof(int) * 2);
	memset(index, 0, sizeof(int) * 2);
	int sum = 0;
	int i = 0, j = numsSize - 1;
	while (i < j)
	{
		sum = tmp[i] + tmp[j];
		if (sum == target)
		{
			break;
		}
		else
		{
			sum>target ? j-- : i++;
		}
	}

	if (i >= j)
	{
		return NULL;
	}

	int m, n;
	for (m = 0; m<numsSize; m++)
	{
		if (nums[m] == tmp[i])
		{
			index[0] = m + 1;
			break;
		}
	}
	for (n = 0; n<numsSize; n++)
	{
		if ((nums[n] == tmp[j]) && (index[0] != (n + 1)))
		{
			index[1] = n + 1;
			break;
		}
	}
	if (index[0]>index[1])
	{
		int ind = index[0];
		index[0] = index[1];
		index[1] = ind;
	}
	
	free(tmp);

	return index;
}