zoukankan      html  css  js  c++  java
  • Leetcode: House Robber

    You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.
    
    Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

    This particular problem and most of others can be approached using the following sequence: https://leetcode.com/problems/house-robber/discuss/156523/From-good-to-great.-How-to-approach-most-of-DP-problems. 

    1. Find recursive relation
    2. Recursive (top-down)
    3. Recursive + memo (top-down)
    4. Iterative + memo (bottom-up)
    5. Iterative + N variables (bottom-up)

    1. Find recursive relation

      rob(i) = max(rob(i-2) + cur, rob(i-1))

    2. Recursive (top-down)

    1 public int rob(int[] nums) {
    2     return rob(nums, nums.length - 1);
    3 }
    4 private int rob(int[] nums, int i) {
    5     if (i < 0) {
    6         return 0;
    7     }
    8     return Math.max(rob(nums, i - 2) + nums[i], rob(nums, i - 1));
    9 }

    3.  Recursive + memo (top-down).

     1 int[] memo;
     2 public int rob(int[] nums) {
     3     memo = new int[nums.length + 1];
     4     Arrays.fill(memo, -1);
     5     return rob(nums, nums.length - 1);
     6 }
     7 
     8 private int rob(int[] nums, int i) {
     9     if (i < 0) {
    10         return 0;
    11     }
    12     if (memo[i] >= 0) {
    13         return memo[i];
    14     }
    15     int result = Math.max(rob(nums, i - 2) + nums[i], rob(nums, i - 1));
    16     memo[i] = result;
    17     return result;
    18 }

    4.  Iterative + memo (bottom-up)

     1 public int rob(int[] nums) {
     2     if (nums.length == 0) return 0;
     3     int[] memo = new int[nums.length + 1];
     4     memo[0] = 0;
     5     memo[1] = nums[0];
     6     for (int i = 1; i < nums.length; i++) {
     7         int val = nums[i];
     8         memo[i+1] = Math.max(memo[i], memo[i-1] + val);
     9     }
    10     return memo[nums.length];
    11 }

    5. Iterative + 2 variables (bottom-up)

     1 /* the order is: prev2, prev1, num  */
     2 public int rob(int[] nums) {
     3     if (nums.length == 0) return 0;
     4     int prev1 = 0;
     5     int prev2 = 0;
     6     for (int num : nums) {
     7         int tmp = prev1;
     8         prev1 = Math.max(prev2 + num, prev1);
     9         prev2 = tmp;
    10     }
    11     return prev1;
    12 }
  • 相关阅读:
    socket.io建立长连接
    编译性语言、解释性语言和脚本语言的区别
    回调函数
    jquery文档处理
    css--Bootstrap框架
    css--960框架
    html基础
    真的理解闭包了吗,用闭包的时候应该注意什么?
    初识reactJs 相关
    简单方便的div垂直居中。
  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/4418967.html
Copyright © 2011-2022 走看看