原文链接: https://leetcode-cn.com/problems/plates-between-candles

英文原文

There is a long table with a line of plates and candles arranged on top of it. You are given a 0-indexed string s consisting of characters '*' and '|' only, where a '*' represents a plate and a '|' represents a candle.

You are also given a 0-indexed 2D integer array queries where queries[i] = [lefti, righti] denotes the substring s[lefti...righti] (inclusive). For each query, you need to find the number of plates between candles that are in the substring. A plate is considered between candles if there is at least one candle to its left and at least one candle to its right in the substring.

• For example, s = "||**||**|*", and a query [3, 8] denotes the substring "*||**|". The number of plates between candles in this substring is 2, as each of the two plates has at least one candle in the substring to its left and right.

Return an integer array answer where answer[i] is the answer to the ith query.

 

Example 1:

Input: s = "**|**|***|", queries = [[2,5],[5,9]]
Output: [2,3]
Explanation:
- queries[0] has two plates between candles.
- queries[1] has three plates between candles.

Example 2:

Input: s = "***|**|*****|**||**|*", queries = [[1,17],[4,5],[14,17],[5,11],[15,16]]
Output: [9,0,0,0,0]
Explanation:
- queries[0] has nine plates between candles.
- The other queries have zero plates between candles.

 

Constraints:

• 3 <= s.length <= 105
• s consists of '*' and '|' characters.
• 1 <= queries.length <= 105
• queries[i].length == 2
• 0 <= lefti <= righti < s.length

中文题目

给你一个长桌子，桌子上盘子和蜡烛排成一列。给你一个下标从 0 开始的字符串 s ，它只包含字符 '*' 和 '|' ，其中 '*' 表示一个 盘子 ，'|' 表示一支 蜡烛 。

同时给你一个下标从 0 开始的二维整数数组 queries ，其中 queries[i] = [lefti, righti] 表示 子字符串 s[lefti...righti] （包含左右端点的字符）。对于每个查询，你需要找到 子字符串中 在 两支蜡烛之间 的盘子的 数目 。如果一个盘子在 子字符串中 左边和右边 都 至少有一支蜡烛，那么这个盘子满足在 两支蜡烛之间 。

• 比方说，s = "||**||**|*" ，查询 [3, 8] ，表示的是子字符串 "*||**|" 。子字符串中在两支蜡烛之间的盘子数目为 2 ，子字符串中右边两个盘子在它们左边和右边 都 至少有一支蜡烛。

请你返回一个整数数组 answer ，其中 answer[i] 是第 i 个查询的答案。

 

示例 1:

输入：s = "**|**|***|", queries = [[2,5],[5,9]]
输出：[2,3]
解释：
- queries[0] 有两个盘子在蜡烛之间。
- queries[1] 有三个盘子在蜡烛之间。

示例 2:

输入：s = "***|**|*****|**||**|*", queries = [[1,17],[4,5],[14,17],[5,11],[15,16]]
输出：[9,0,0,0,0]
解释：
- queries[0] 有 9 个盘子在蜡烛之间。
- 另一个查询没有盘子在蜡烛之间。

 

提示：

• 3 <= s.length <= 105
• s 只包含字符 '*' 和 '|' 。
• 1 <= queries.length <= 105
• queries[i].length == 2
• 0 <= lefti <= righti < s.length

通过代码

高赞题解

func platesBetweenCandles(s string, queries [][]int) []int {
	n := len(s)
	sum := make([]int, n+1) // sum[i] 表示 s[:i] 中盘子的个数
	left := make([]int, n) // left[i] 表示 i 左侧最近蜡烛位置
	p := -1
	for i, b := range s {
		sum[i+1] = sum[i]
		if b == '|' {
			p = i
		} else {
			sum[i+1]++
		}
		left[i] = p
	}

	right := make([]int, n) // right[i] 表示 i 右侧最近蜡烛位置
	for i, p := n-1, n; i >= 0; i-- {
		if s[i] == '|' {
			p = i
		}
		right[i] = p
	}

	ans := make([]int, len(queries))
	for i, q := range queries {
		l, r := right[q[0]], left[q[1]] // 用最近蜡烛位置来代替查询的范围，从而去掉不符合要求的盘子
		if l < r {
			ans[i] = sum[r] - sum[l]
		}
	}
	return ans
}

统计信息

通过次数	提交次数	AC比率
2030	5254	38.6%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言