力扣刷题记录

1. 两数之和/** * 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(2 * sizeof(int));//malloced *returnSize 2; // 设置返回数组的长度 for(int i 0;i(numsSize);i) { for(int ji1;j(numsSize);j) { if((i!j) (nums[i]nums[j]target)) { result[0]i; result[1]j; return result; //找到答案 } } } free(result); *returnSize 0; return NULL; }2. 两数相加# Definition for singly-linked list. # class ListNode: # def __init__(self, val0, nextNone): # self.val val # self.next next class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) - Optional[ListNode]: dummy ListNode(0) curr dummy carry 0 #进位 while l1 or l2 or carry: val1 l1.val if l1 else 0 val2 l2.val if l2 else 0 total val1 val2 carry carry total // 10; digit total % 10; curr.next ListNode(digit) curr curr.next if l1: l1 l1.next if l2: l2 l2.next return dummy.next