Problem 1: two sum
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2
.gitignore
vendored
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2
.gitignore
vendored
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*.o
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bin
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problems/1-two-sum/Makefile
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problems/1-two-sum/Makefile
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all: lib.o main.o
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gcc -Wall -o bin lib.o main.o
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lib.o: lib.c
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gcc -c -Wall lib.c
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main.o: main.c
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gcc -c -Wall main.c
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run:
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./bin
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clean:
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rm *.o bin
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problems/1-two-sum/lib.c
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problems/1-two-sum/lib.c
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#include <stdlib.h>
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#include "lib.h"
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int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
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for (int i=0; i<numsSize-1; i++) {
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for (int j=i+1; j<numsSize; j++) {
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if (nums[i] + nums[j] == target) {
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int* res = malloc(2 * sizeof(int));
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res[0] = i;
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res[1] = j;
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*returnSize = 2;
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return res;
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}
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}
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}
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*returnSize = 0;
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return NULL;
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}
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2
problems/1-two-sum/lib.h
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problems/1-two-sum/lib.h
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#pragma once
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int* twoSum(int* nums, int numsSize, int target, int* returnSize);
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problems/1-two-sum/main.c
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problems/1-two-sum/main.c
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#include "stdio.h"
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#include "stdlib.h"
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#include "lib.h"
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int main() {
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int returnSize = -1;
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int nums1[] = {2, 7, 11, 15};
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int target1 = 9;
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int* res1 = twoSum(nums1, 4, target1, &returnSize);
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if (returnSize == 2 && res1[0] == 0 && res1[1] == 1) {
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printf("Test 1: passed\n");
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}
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else {
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printf("Test 1: failed\n");
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}
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if (returnSize > 0) {
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free(res1);
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}
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int nums2[] = {3, 2, 4};
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int target2 = 6;
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int* res2 = twoSum(nums2, 3, target2, &returnSize);
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if (returnSize == 2 && res2[0] == 1 && res2[1] == 2) {
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printf("Test 2: passed\n");
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}
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else {
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printf("Test 2: failed\n");
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}
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if (returnSize > 0) {
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free(res2);
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}
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return 0;
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}
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9
problems/1-two-sum/readme.md
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problems/1-two-sum/readme.md
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# 1. Two sum
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Given an array of integers nums and an integer target,
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return indices of the two numbers such that they add up to target.
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You may assume that each input would have exactly one solution,
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and you may not use the same element twice.
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You can return the answer in any order.
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