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leetcode 顺序刷题,详细通俗题解,with JAVA

Home Page: https://leetcode.wang

Shell 60.44% CSS 39.56%
leetcode leetcode-java

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leetcode's Issues

公式括号问题

这篇文章

此时 3 位置的数值对应为数组下标是 0 的值,1 位置的值对应数组下标是 5 的值。任意位置的对应规则是什么呢?0 -> 4, 1 - > 5,4 ->1,就是就是 (位置 + 偏移 % 数组的长度)。这里就是加上 4 模 7。

其中,(位置 + 偏移 % 数组的长度)应该改成(位置 + 偏移)% 数组的长度

leetCode-2-Add-Two-Numbers 扩展

/*
链表反序存储时,是不是可以使用堆栈?
*/

type stack []int

func (s *stack) push(item int) {
	(*s) = append((*s), item)
}

func (s *stack) pop() int {
	item := (*s)[len(*s)-1]
	(*s) = (*s)[:len(*s)-1]
	return item
}

func (s stack) depth() int {
	return len(s)
}

func addTwoReverseNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
	sumStack := make(stack, 0)
	carryStack := make(stack, 0)
	carry := 0
	for l1 != nil || l2 != nil {
		l1val := 0
		l2val := 0
		if l1 != nil {
			l1val = l1.Val
			l1 = l1.Next
		}
		if l2 != nil {
			l2val = l2.Val
			l2 = l2.Next
		}
		sum := l1val + l2val
		carry = sum / 10
		sumStack.push(sum % 10)
		carryStack.push(carry)
	}
	sumList := new(ListNode)

	isFirst := true
	carry = 0
	for sumStack.depth() != 0 || carryStack.depth() != 0 || carry != 0 {
		curSum := 0
		curCarry := 0
		curNode := new(ListNode)
		if sumStack.depth() != 0 {
			curSum = sumStack.pop()
		}
		if !isFirst && carryStack.depth() != 0 {
			curCarry = carryStack.pop()
		}
		isFirst = false
		curVal := (curSum + curCarry + carry)
		carry = curVal / 10
		curNode.Val = curVal % 10
		curNode.Next = sumList.Next
		sumList.Next = curNode
	}
	return sumList.Next
}

49 題的內容建議

也不算 issue 就是

不確定文本寫成時是否 AC,但目前解法一會超時,建議注釋一下

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