【436】Solution for LeetCode Problems

Coding everyday. ^_^

1. Two Sum

• 重点知识：指针可以存储数值，通过 malloc 新建数组
• int* returnSize：Size of the return array. Store the value in a pointer, say 2.
  *returnSize = 2
• My solution:
• ?

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  /**  * Note: The returned array must be malloced, assume caller calls free().  */ int* twoSum(int* nums, int numsSize, int target, int* returnSize){     *returnSize = 2;     int* returnArray = malloc(sizeof(int)*(*returnSize));     for (int i = 0; i < numsSize-1; i++) {         for (int j = i+1; j < numsSize; j++) {             if (nums[i] + nums[j] == target) {                 returnArray[0] = i;                 returnArray[1] = j;                 return returnArray;             }         }     }     returnArray[0] = -1;     returnArray[1] = -1;     return returnArray; }

   

2. Add Two Numbers

• 重点知识：不能通过数字来计算，考虑进位，考虑链表遍历和插入节点，通过 malloc 新建节点
• Don't use integer to calculate in this problem since the numbers are very long.
• Need to consider carry bit.
• This linked list's head is the first node.
• My solution:
• ?

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  /**  * Definition for singly-linked list.  * struct ListNode {  *     int val;  *     struct ListNode *next;  * };  */     struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){     struct ListNode* p1;     struct ListNode* p2;     struct ListNode* l3 = NULL;     struct ListNode* p3 = NULL;     p1 = l1;     p2 = l2;           int carry_bit = 0;     while (p1 != NULL || p2 != NULL || carry_bit == 1) {         int num1;         int num2;         int num3;         if (p1 == NULL && p2 == NULL && carry_bit == 1) {             num3 = 1;             carry_bit = 0;         }         else {             if (p1 == NULL) {                 num1 = 0;             }             else {                 num1 = p1->val;                 p1 = p1->next;             }             if (p2 == NULL) {                 num2 = 0;             }             else {                 num2 = p2->val;                 p2 = p2->next;             }                           num3 = num1 + num2 + carry_bit;             if (num3 >= 10) {                 carry_bit = 1;                 num3 = num3 - 10;             }             else {                 carry_bit = 0;             }         }           struct ListNode* tmp;         tmp = malloc(sizeof(struct ListNode));         if (tmp == NULL) {             fprintf(stderr, "Out of memory.\n");             exit(1);         }         tmp->val = num3;         tmp->next = NULL;                   if (p3 == NULL) {             p3 = tmp;             l3 = p3;         }         else {             p3->next = tmp;             p3 = p3->next;         }     }           return l3; }

 

3. Longest Substring Without Repeating Characters

• 重点知识：多层遍历，时间复杂度过高
• My solution:
• ?

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  int lengthOfLongestSubstring(char * s){     int length = strlen(s);     if (length == 1){         return 1;     }     int max = 0;     for (int i = 0; i < length; i++) {         int flag = 1;         for (int j = i + 1; j < length & flag; j++) {             for (int k = j - 1; k >= i; k--) {                 if (s[j] == s[k]) {                     int tmp = j - i;                     if (max < tmp) {                         max = tmp;                     }                     i = k;                     flag = 0;                     break;                 }                 if (k == i) {                     int tmp = j - i + 1;                     if (max < tmp) {                         max = tmp;                     }                 }             }         }     }     return max; }