給定一個(gè)整數(shù)數(shù)組和一個(gè)目標(biāo)值，找出數(shù)組中和為目標(biāo)值的兩個(gè)數(shù)。
你可以假設(shè)每個(gè)輸入只對(duì)應(yīng)一種答案，且同樣的元素不能被重復(fù)利用赔退。
示例:
給定 nums = [2, 7, 11, 15], target = 9
因?yàn)?nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]
C語言
#include <stdio.h>
#include <stdlib.h>
int* twoSum(int* nums, int numsSize, int target) {
int min = 2147483647;
int i = 0;
for (i = 0; i < numsSize; i++) {
if (nums[i] < min)
min = nums[i];
}
int max = target - min;
int len = max - min + 1; //確定hash長(zhǎng)度
int *table = (int*)malloc(len*sizeof(int));
int *indice = (int*)malloc(2*sizeof(int));
for (i = 0; i < len; i++) {
table[i] = -1; //hash初值
}
for (i = 0; i < numsSize; i++) {
if (nums[i]-min < len) {
if (table[target-nums[i]-min] != -1) { //滿足相加為target
indice[0] = table[target-nums[i] - min];
indice[1] = i;
return indice;
}
table[nums[i]-min] = i;
}
}
free(table);
return indice;
}
int main() {
int a[4] = {2,7,11,13};
int tag = 9;
int *num = twoSum(a,4,tag);
printf("%d-%d",num[0],num[1]);
return 0;
}
