LeetCode Two-Sum
Questions:Two-Sum
My Answer:
/*
    Function: Copynums
    Explanation:
        The function is used to judging whether a number is in nums or not.
 */
int* copynums(int* nums, int numsSize, int maxn, int minn)
{
    int i;
    
    int* cpnums;
    
    cpnums = (int*)malloc((maxn-minn+1) * sizeof(int));
    
    for (i = 0; i < maxn-minn; i++)
    {
        *(cpnums+i) = 0;
    }
    
    for (i = 0; i < numsSize; i++)
    {
        cpnums[nums[i]-minn] = 1;
    }
    
    return cpnums;
}
int* twoSum(int* nums, int numsSize, int target)
{
    int* ans;
    
    ans = (int*)malloc(5*sizeof(int));
    
    int i, j;
    
    int maxn = 0, minn = 0;
    int pla1 = 0, pla2 = 0;
    
    maxn = nums[0];
    minn = nums[0];
    
    for (i = 1; i < numsSize; i++)
    {
        if (nums[i] > maxn) maxn = nums[i];
        if (nums[i] < minn) minn = nums[i];
    }
    
    int* cpnums = (int*)malloc((maxn-minn+1) * sizeof(int));
    
    cpnums = copynums(nums, numsSize, maxn, minn);
    
    for (i = 0; i < numsSize; i++)
    {
        int wanna = target - nums[i];
        
        bool flag = false;
        
        if (cpnums[wanna-minn] == 1)            // the desire number is in nums
        {
            for (j = 0; j < numsSize; j++)      // find its pla
            {
                if (nums[j] == wanna && i != j)
                {
                    flag = true;
                    
                    pla1 = i;
                    pla2 = j;
                    
                    break;
                }
            }
        }
        
        if (flag) break;
    }
    
    ans[0] = pla1;
    ans[1] = pla2;
    
    return ans;
}
Complexity:O(n^2)
Examples
Examples_A:
Input:
[3,2,4]
6
Except Output:
[1,2]
Examples_B:
Input:
[-1,-2,-3,-4]
-3
Except Output:
[0,1]
Examples_C:
Input:
[-3,2,3,5]
0
Except Output:
[0,2]
Hint
使用malloc()函数进行内存分配时应注意分配内存的大小,如
int* cpnums = (int*)malloc((maxn-minn+1) * sizeof(int));
分配(maxn-minn+1)个int的内存单元大小。
Better Solutions:solutions
2017/1/28
    To improve is to change, to be perfect is to change often.

 
                
            
         
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浙公网安备 33010602011771号