198. House Robber

  • Hardness: \(\color{orange}\textsf{Medium}\)
  • Ralated Topics: Array、Dynamic Programming

一、題目

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 systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night. Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:

  • Input: nums = [1,2,3,1]
  • Output: 4
  • Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). Total amount you can rob = 1 + 3 = 4.

Example 2:

  • Input: nums = [2,7,9,3,1]
  • Output: 12
  • Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1). Total amount you can rob = 2 + 9 + 1 = 12.

Constraints:

  • 1 <= nums.length <= 100
  • 0 <= nums[i] <= 400

二、分析

  • 時間序列型的動態規劃問題。
    • 定義 dp[i][j]：第 i 間房子，j == 0 代表不搶，j == 1 代表搶。
    • 第 i 間房若搶，則前一間房必定不能搶；第 i 間房若不搶，前一間房可搶可不搶：
      • dp[i][0] = max(dp[i-1][1], dp[i-1][0])
      • dp[i][1] = dp[i-1][0] + val[i]
    • 最終的結果是 max(dp[n-1][0], dp[n-1][1])。

三、解題

1. Dynamic Programming

  • Time complexity: \(O(n)\)
  • Space complexity: \(O(n)\)
int rob(vector<int>& nums) {
    int n = nums.size();
    vector<vector<int>> dp(n, vector<int>(2, 0));
    dp[0][1] = nums[0];
    for (int i = 1; i < n; i++) {
        dp[i][0] = max(dp[i-1][1], dp[i-1][0]);
        dp[i][1] = dp[i-1][0] + nums[i];
    }
    return max(dp[n-1][0], dp[n-1][1]);
}

2. Dynamic Programming(SC optimized)

  • Time complexity: \(O(n)\)
  • Space complexity: \(O(1)\)
int rob(vector<int>& nums) {
    int n = nums.size();
    int robbed = nums[0];
    int passed = 0;
    for (int i = 1; i < n; i++) {
        int tmp = robbed;
        robbed = passed + nums[i];
        passed = max(tmp, passed);
    }
    return max(robbed, passed);
}

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