【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:

• /**
   * 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:

• /**
   * 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:

• 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;
  }