leetcode笔记 191. Number of 1 Bits & 198. House Robber_where 198 code number is-优快云博客

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本文提供LeetCode上两道经典题目“Number of 1 Bits”和“House Robber”的详细解析及代码实现。“Number of 1 Bits”要求计算输入整数的二进制表示中1的个数；“House Robber”则考验如何在不触动相邻房屋警报的情况下，最大化抢劫所得金钱。

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leetcode刷题24天
191. Number of 1 Bits
Write a function that takes an unsigned integer and return the number of ‘1’ bits it has (also known as the Hamming weight).

Example 1:

Input: 00000000000000000000000000001011
Output: 3
Explanation: The input binary string 00000000000000000000000000001011 has a total of three ‘1’ bits.
Example 2:

Input: 00000000000000000000000010000000
Output: 1
Explanation: The input binary string 00000000000000000000000010000000 has a total of one ‘1’ bit.
Example 3:

Input: 11111111111111111111111111111101
Output: 31
Explanation: The input binary string 11111111111111111111111111111101 has a total of thirty one ‘1’ bits.

代码：

public class Solution {
    // you need to treat n as an unsigned value
    public int hammingWeight(int n) {
        if(n==0)
        	return 0;
        int res = 0;
        while(n!=0) {
        	if((n & 1) == 1)
        		res++;
        	n = n>>>1;
        }
    
        
        return res;
    }
}

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.

Example 1:

Input: [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: [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.

代码：

class Solution {
    public int rob(int[] nums) {
        if(nums.length==0)
        	return 0;
        int[] dp = new int[nums.length];
        dp[0] = nums[0];
        for(int i=1;i<nums.length;i++) {
        	if(i==1)
        		dp[i] = Math.max(dp[0], nums[i]);
        	else
        		dp[i] = Math.max(dp[i-1], dp[i-2]+nums[i]);
        }
        return dp[nums.length-1];        
    }
}