LeetCode Weekly Contest 144

怼烎@ 2023-06-01 05:48 19阅读 0赞

### 1108. Defanging an IP Address ###

Given a valid (IPv4) IP `address`, return a defanged version of that IP address.

A *defanged IP address* replaces every period `"."` with `"[.]"`.

Example 1:

Input: address = "1.1.1.1"
    Output: "1[.]1[.]1[.]1"

Example 2:

Input: address = "255.100.50.0"
    Output: "255[.]100[.]50[.]0"

Constraints:

*  The given `address` is a valid IPv4 address.

题目大意：给你一个ip地址，任务是将ip地址的"."改成"\[.\]"。

思路：直接模拟就好，或者直接用replace方法。

class Solution {
    	public String defangIPaddr(String address) {
    		int len = address.length();
    		StringBuffer bf = new StringBuffer();
    		for(int i=0; i<len; i++) {
    			char ch = address.charAt(i);
    			if( ch == '.' ) bf.append("[.]");
    			else bf.append(ch);
    		}
    		return bf.toString();
    	}
    }

### 1109. Corporate Flight Bookings ###

There are `n` flights, and they are labeled from `1` to `n`.

We have a list of flight bookings.  The `i`\-th booking `bookings[i] = [i, j, k]` means that we booked `k` seats from flights labeled `i` to `j`inclusive.

Return an array `answer` of length `n`, representing the number of seats booked on each flight in order of their label.

Example 1:

Input: bookings = [[1,2,10],[2,3,20],[2,5,25]], n = 5
    Output: [10,55,45,25,25]

Constraints:

*  `1 <= bookings.length <= 20000`
 *  `1 <= bookings[i][0] <= bookings[i][1] <= n <= 20000`
 *  `1 <= bookings[i][2] <= 10000`

题目大意：给你一个航班预定表，表中第 `i` 条预订记录 `bookings[i] = [i, j, k]` 意味着我们在从 `i` 到 `j` 的每个航班上预订了 `k` 个座位。返回一个长度为 `n` 的数组 `answer`，按航班编号顺序返回每个航班上预订的座位数。

思路：直接看就是桶排，只不过长度太长，直接暴力肯定会时间超限。这时就需要时间优化了，看题目是i到j连续的，第一反应的优化方法就是前缀和。

class Solution {
    	public int[] corpFlightBookings(int[][] bookings, int n) {
    		int[] a = new int[n+1];
    		for(int[] b : bookings){
    			a[b[0]-1] += b[2];
    			a[b[1]] -= b[2];
    		}
    		for(int i = 0;i < n;i++){
    			a[i+1] += a[i];
    		}
    		return Arrays.copyOf(a, n);
    	}
    }

转载于:https://www.cnblogs.com/Asimple/p/11258509.html