关于leetcode:Leetcode刷题笔记第1天

java程序猿转go语言，通过leetcode刷题来相熟go语言，实现语言根底和语法根底的相熟

1.两数之和

暴力破解:

func twoSum(nums []int, target int) []int {
    for i := 0; i < len(nums) - 1; i++ {
        for j := i + 1; j < len(nums); j++ {
            if (nums[j] + nums[i] == target) {
                return []int {i, j};
            }
        }
    }
    return nil
}

哈希表：

func twoSum(nums []int, target int) []int {
    hashmap := make(map[int] int)
    for index, value := range nums {
        if v, ok := hashmap[value]; ok {
            return []int{index, v}
        }
        hashmap[target - value] = index
    }
    return nil
}

2.两数相加

递归版本（链表或者树的问题都要尽量想一个递归的）

 func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
    return addTwoNumbersHelp(l1, l2, 0);
}
func addTwoNumbersHelp(l1 *ListNode, l2 *ListNode, add int) *ListNode {
   if l1 == nil && l2 == nil && add == 0 {
      return nil
 }
   if (l1 != nil) {
      add = l1.Val + add;
      l1 = l1.Next
 }
   if (l2 != nil) {
      add = l2.Val + add;
      l2 = l2.Next
 }
   node := ListNode{
      add % 10,
      addTwoNumbersHelp(l1, l2, add / 10),
   }
   return &node;
}

非递归版本

func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
    head := ListNode{};
    tail := &head;
    add := 0;
    for l1 != nil && l2 != nil {
        add += l1.Val + l2.Val;
        l1 = l1.Next;
        l2 = l2.Next;
        cur := ListNode{
            Val: add % 10,
            Next: nil,
        }
        tail.Next = &cur;
        tail = &cur;
        add = add / 10;
    }
    for l1 != nil {
        add += l1.Val;
        l1 = l1.Next;
        cur := ListNode{
            Val: add % 10,
            Next: nil,
        }
        tail.Next = &cur;
        tail = &cur;
        add = add / 10;
    }
    for l2 != nil {
        add += l2.Val;
        l2 = l2.Next;
        cur := ListNode{
            Val: add % 10,
            Next: nil,
        }
        tail.Next = &cur;
        tail = &cur;
        add = add / 10;
    }
    if add != 0 {
        tail.Next = &ListNode{1,nil}
    }
    return head.Next;
}