给定一个整数数组 nums 和一个目标值 target,请你在该数组中找出和为目标值的那 两个 整数,并返回他们的数组下标。
你可以假设每种输入只会对应一个答案。但是,数组中同一个元素不能使用两遍。
方法1:双循环
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* twoSum(int* nums, int numsSize, int target, int* returnSize){
int *result = (int *)malloc(sizeof(int) * 2);
for (int i=0; i<numsSize -1; i++)
{
for (int j=i+1; j<numsSize; j++)
{
if ( nums[i] + nums[j] == target)
{
result[0] = i;
result[1] = j;
*returnSize = 2;
return result;
}
}
}
return 0;
}
方法2:手动编写简单散列表
看题解后去学习了一下,这里用的取余方法
hash(key) = key%maxsize;
一次循环,每次把下标填到hash表中,用target-nums[i]寻找另外一个数,
如果为-1表示不存在,否则存在两个数的和为target。
扫描二维码关注公众号,回复:
11936260 查看本文章
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* twoSum(int* nums, int numsSize, int target, int* returnSize){
static int result[2] = {0};
static int maxsize=2000;
int hash[maxsize];
memset(hash,-1,sizeof(hash));
for(int i=0; i<numsSize; i++)
{
if(hash[ (target - nums[i] + maxsize) % maxsize ] != -1 )
{
result[0] = hash[(target - nums[i] + maxsize) % maxsize];
result[1] = i;
*returnSize = 2;
return result;
}
else
{
hash[ (nums[i] + maxsize) % maxsize ] = i;
}
}
*returnSize = 0;
return 0;
}
Python 3:
Hash:
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
MaxIndex = 2000;
Hash = [-1 for i in range(MaxIndex)]
result = [None for i in range(2)]
lenOfList = len(nums)
for i in range(lenOfList):
if(Hash[(target - nums[i] + MaxIndex) % MaxIndex] != -1):
result[0] = Hash[(target - nums[i] + MaxIndex) % MaxIndex]
result[1] = i
break
else:
Hash[(nums[i] + MaxIndex) % MaxIndex] = i
return result
2、字典
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
Hash={} # 字典
for i,element in enumerate(nums):
if (target - element) in Hash: #查找元素
return [Hash[target - element], i]
else:
Hash[element] = i # 记录位置 样式: element:i
return None