zoukankan      html  css  js  c++  java
  • Divide and Conquer

    1

    2 218 The Skyline Problem     最大堆   遍历节点

        public List<int[]> getSkyline(int[][] buildings) {
            List<int[]> res = new ArrayList<>();
            List<int[]> point = new ArrayList<>();
            for (int i = 0; i < buildings.length; i++)
            {
                point.add(new int[]{buildings[i][0], buildings[i][2]});
                point.add(new int[]{buildings[i][1], -buildings[i][2]});
            }
            Collections.sort(point, (a,b)-> {
                if (a[0] != b[0]) return a[0] - b[0];
                return b[1] - a[1];});
            PriorityQueue<Integer> maxheap = new PriorityQueue<>((a,b)->(b - a));
            int cur = 0, pre = 0;
            for (int i = 0; i < point.size(); i++)
            {
                int[] b = point.get(i);
                if (b[1] > 0)
                {
                    maxheap.add(b[1]);
                    cur = maxheap.peek();
                }
                else
                {
                    maxheap.remove(-b[1]);
                    cur = maxheap.peek() == null ? 0 : maxheap.peek();
                }
                if (pre != cur)
                {
                    res.add(new int[]{b[0], cur});
                    pre = cur;
                }
            }
            return res;
        }
    View Code

     3  241 Different Ways to Add Parentheses   找符号,分开

        public List<Integer> diffWaysToCompute(String input) {
            List<Integer> res = new LinkedList<>();
            for (int i = 0; i < input.length(); i++)
            {
                char c = input.charAt(i);
                if (c == '+' || c == '-' || c == '*')
                {
                    String a = input.substring(0, i);
                    String b = input.substring(i+1);
                    List<Integer> a1 = diffWaysToCompute(a);
                    List<Integer> b1 = diffWaysToCompute(b);
                    for (int x : a1)
                    {
                        for (int y : b1)
                        {
                            if (c == '+')
                            {
                                res.add(x + y);
                            }
                            else if (c == '-')
                            {
                                res.add(x - y);
                            }
                            else if (c == '*')
                            {
                                res.add(x * y);
                            }
                        }
                    }
                }
            }
            if (res.size() == 0) res.add(Integer.valueOf(input));
            return res;
        }
    View Code
  • 相关阅读:
    局域网中电脑之间无法ping通问题
    Python 字典dict操作定义
    set集合
    Python 元组遍历排序操作方法
    Python List 列表list()方法
    set函数&操作
    dict函数
    文件操作
    list函数
    字符串格式化format使用
  • 原文地址:https://www.cnblogs.com/whesuanfa/p/6817014.html
Copyright © 2011-2022 走看看