題目
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
解答方法一：雙重循環嘗試， 複雜度O(n^2)
#include<stdio.h>
#include<stdlib.h>
int *twoSum(int numbers[], int n, int target)
{
if ((numbers == NULL)|(n < 2))
{
return NULL;
}
int index1;
int index2;
int* ret = (int*)malloc(sizeof(int)*2);
for (index1 = 0; index1 < n; ++index1)
{
for (index2 = index1 + 1; index2 < n; ++index2)
{
int sum_tmp = numbers[index1] + numbers[index2];
if (sum_tmp == target)
{
ret[0] = index1 + 1;
ret[1] = index2 + 1;
return ret;
}
}
}
return NULL;
}
int main(int argc, char **argv)
{
int array[] = {2, 7, 11, 15};
size_t n = sizeof(array)/sizeof(int);
int target = 9;
int* result = twoSum(array, (int)n, target);
printf("The index1: %d; the index2: %d\n", result[0], result[1]);
free(result);//memory deallocation
}
點評：
解法非常直觀， 但是複雜度O(n^2)，所以效率很差，在leetcode上無法通過： Submission Result:
Time Limit Exceeded
解答方法二：用hash表儲存每個數的下標，增加空間使用來加速查找至O(n)
分析： 之前O(n^2)的原因是我們做了太多次無謂的運算。事實上，既然我們知道了和和一個參數，只要判斷下他們的差（和 - 參數一）存不存在即可。
而用哈希表的話，這個判斷存不存在只有O(1)。
下面的代碼我是自己實現的hash，所以比C++裏用STL代碼長很多，但是效率分要比C++高（C++代碼有個網友貼了，44ms），而我這個C代碼是13ms，大概也是哈希表解法的上限了。
C code
有兩個函數 twoSum是解法一； twoSum_hash是解法二
#include<stdlib.h>
#define PRIME 271
int *twoSum(int numbers[], int n, int target)
{
if ((numbers == NULL)|(n < 2))
{
return NULL;
}
int index1;
int index2;
int* ret = (int*)malloc(sizeof(int)*2);
if (NULL == ret)
{
printf("memory allocation for ret failed!\n");
return NULL;
}
for (index1 = 0; index1 < n; ++index1)
{
for (index2 = index1 + 1; index2 < n; ++index2)
{
int sum_tmp = numbers[index1] + numbers[index2];
if (sum_tmp == target)
{
ret[0] = index1 + 1;
ret[1] = index2 + 1;
return ret;
}
}
}
return NULL;
}
//second method
typedef struct linkedListNode* plistNode;
typedef struct linkedListNode
{
int index;
plistNode next;
}listNode;
void hashTableFree(plistNode hashTable, int primeNumber)
{
int index = 0;
for (index = 0; index < primeNumber; ++index)
{
plistNode tail = hashTable[index].next;
while(NULL != tail)
{
plistNode tmp = tail;
tail = tail->next;
free(tmp);
}
}
free(hashTable);
}
int *twoSum_hash(int numbers[], int n, int target)
{
if ((numbers == NULL)|(n < 2))
{
return NULL;
}
int index = 0;
//create a hash table
int primeNumber = PRIME;
plistNode hashTable = (plistNode)malloc(sizeof(listNode)*primeNumber);
if (NULL == hashTable)
{
printf("memory allocation for hashTable failed!\n");
return NULL;
}
//initialise the hash table
for (index = 0; index < primeNumber; ++index)
{
hashTable[index].index = -1;
hashTable[index].next = NULL;
}
//assign numbers to the hash table
for (index = 0; index < n; ++index)
{
int bias = abs(numbers[index]%primeNumber);
if (-1 == hashTable[bias].index)
{
hashTable[bias].index = index;
}else //occupied
{
plistNode tail = hashTable[bias].next;
tail = (plistNode )malloc(sizeof(listNode));
if (NULL == tail)
{
printf("memory allocation for tail failed!\n");
hashTableFree(hashTable, primeNumber);
return NULL;
}
tail->index = index;
tail->next = hashTable[bias].next;
hashTable[bias].next = tail;
}
}
//searching
int* ret = (int*)malloc(sizeof(int)*2);
for (index = 0; index < n; ++index )
{
int bias = abs((target - numbers[index])%primeNumber);
if (-1 == hashTable[bias].index) // this bias doesn't exist
{
continue;
}
if ((numbers[hashTable[bias].index] == (target - numbers[index])) & (hashTable[bias].index != index))
{
ret[0] = index + 1;
ret[1] = hashTable[bias].index + 1;
hashTableFree(hashTable, primeNumber);
return ret;
}else
{
plistNode tail = hashTable[bias].next;
while (NULL != tail)
{
if (numbers[tail->index] == (target - numbers[index]))
{
ret[0] = index + 1;
ret[1] = tail->index + 1;
hashTableFree(hashTable, primeNumber);
return ret;
}else
{
tail = tail->next;
}
}
}
}
hashTableFree(hashTable, primeNumber);
return NULL;
}
int main(int argc, char **argv)
{
//int array[] = {438,507,629,255,813,423,536,428,340,767,208,808,882,142,835,423,331,545,627,750,397,675,662,92,465,627,15,522,395,727,561,73,570,606,826,651,743,214,881,685,820,326,653,334,698,604,938,260,51,597,291,855,427,117,943,142,166,439,833,4,180,10,531,350,785,989,607,303,554,764,769,451,654,752,15,90,505,159,1,516,801,938,442,543,761,548,523,766,445,696,217,352,333,383,868,764,556,943,280,140,627,870,635,753,743,978,611,326,405,143,564,256,304,913,570,331,340,222,952,959,535,113,148,125,874,354,984,753,423,448,235,621,796,355,64,682,326,500,609,293,566,974,808,568,729,173,735,764,987,588,227,961,621,340,245,570,640,814,635,482,520,563,695,399,95,92,813,135,342,513,410,943,64,458,801,835,977,932,838,604,500,266,395,108,788,161,769,662,697,167,143,383,880,19,758,552,396,226,548,560,916,766,568,192,100,734,639,288,187,465,345,535,293,130,488,172,108,313,800,662,644,758,843,953,73,543,630,37,711,372,372,410,60,973,750,833,827,572,31,157,473,529,410,650,930,635,253,203,159,431,848,969,982,229,504,617,273,723,854,572,879,382,18,418,990,321,871,544,662,900,340,358,788,998,376,87,520,543,717,67,713,419,108,796,143,342,896,526,926,362,23,63,426,514,526,499,857,87,499,635,496,69,558,635,1,635,125,611,459,11,14,305,969,225,677,354,90,509,317,895,617,798,210,830,63,488,139,500,744,564,295,782,214,569,863,835,14,70,687,321,130,933,822,761,985,637,995,604,932,54,51,170,951,774,117,117,863,940,599,820,473,350,572,776,198,958,1,281,539,874,38,205,591,901,588,743,926,270,675,135,208,588,438,444,752,172,709,191,199,494,743,854,362,340,170,909,434,258,621,983,119,37,281,987,53,486,709,824,817,590,632,798,207,691,592,358,92,316,864,345,488,408,172,392,402,188,880,336,350,739,301,953,314,414,261,701,431,984,350,800,476,370,952,580,585,335,14,486,448,710,637,134,665,166,631,526,877,940,492,855,563,417,517,543,496,979,802,100,100,854,706,350,796,140,46,434,811,543,743,614,838,803,583,280,748,124,570,134,18,282,879,324,142,904,861,884,974,630,319,340,130,954,130,199,681,465,295,597,492,893,552,920,370,221,293,26,632,695,967,372,321,723,907,701,733,520,482,927,11,606,466,404,529,402,166,646,60,802,101,822,93,12,336,136,650,539,640,791,108,997,268,833,562,132,423,879,884,362,374,522,188,496,472,531,434,942,46,528,423,504,664,867,467,685,899,884,746,548,937,296,436,295,486,722,288,465,89,736,900,174,497,37,84,817,411,486,387,949,893,187,776,771,617,175,516,705,296,452,268,6,922,402,991,723,609,98,347,588,554,80,169,197,931,19,597,888,164,347,27,916,378,630,321,644,156,362,4,705,164,465,543,806,282,827,448,670,950,458,388,239,744,507,516,414,459,934,130,924,440,166,788,343,379,884,758,808,912,187,706,419,88,355,996,331,777,500,863,796,627,124,535,561,466,370,101,383,813,44};
int array[] = {3,2,4};
size_t n = sizeof(array)/sizeof(int);
int target = 6;
int* result = twoSum_hash(array, (int)n, target);
if (NULL != result)
{
printf("The index1: %d; the index2: %d\n", result[0], result[1]);
printf("values are %d and %d.", array[result[0]-1], array[result[1]-1]);
free(result);//memory deallocation
}else
{
printf("result is not available!\n");
}
}