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 jinclusive.

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);
	}
}