1.题目:
2.题解:
2.1为什么用滑动窗口优化:
因为元素都是大于0的
所以:当找到大于等于x的值时,right可以不用返回
两个指针都往后走;因此可以使用滑动窗口优化暴力解法
2.2:滑动窗口具体使用步骤:
1.进窗口:sum += array[right];
2.判断:sum >= x 时出窗口
灵活更新结果(满足结果后) right-left+1<retlen
3.出窗口:sum -= array[left];
图解:
代码:这里注意使用了一个读取模板,不让Scanner输入会超时
javaimport java.util.*; import java.io.*; public class Main { public static void main(String[] args) throws IOException { Read in = new Read(); int n = in.nextInt(), x = in.nextInt(); int[] array = new int[n+1];//注意下标从1开始 for(int i = 1; i <= n; i++){ array[i] = in.nextInt(); } int left = 1; int right = 1; int sum = 0; int retlen = n; int retLeft = -1; int retRight = -1; while(right <= n){ //进窗口 sum += array[right]; //判断 while(sum >= x){ //更新结果 if(right-left+1 < retlen){ retLeft = left; retRight = right; retlen = right-left+1;//更新以便于找出最小值 } //出窗口 sum -= array[left++]; } right++; } System.out.print(retLeft +" " +retRight); } } class Read // 自定义快速读入 { //字符串截取 StringTokenizer st = new StringTokenizer(""); //1.字节流->字符流:new InputStreamReader(System.in) //2.带内存缓冲区的字符流:BufferedReader BufferedReader bf = new BufferedReader(new InputStreamReader(System.in)); String next() throws IOException { while(!st.hasMoreTokens()) { ///读内存缓冲区里的一行数据:bf.readLine() st = new StringTokenizer(bf.readLine()); } //获取每一个截取的字符串 return st.nextToken(); } //转化为自己想要的类型 String nextLine() throws IOException { return bf.readLine(); } int nextInt() throws IOException { return Integer.parseInt(next()); } long nextLong() throws IOException { return Long.parseLong(next()); } double nextDouble() throws IOException { return Double.parseDouble(next()); } }
滑动窗口->dd爱框框
robin_suli2024-10-06 21:04