Leetcode-477. Total Hamming Distance

最新推荐文章于 2021-04-10 14:46:54 发布

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最新推荐文章于 2021-04-10 14:46:54 发布

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分类专栏：算法文章标签： java leetcode 算法

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本文解析了一道经典的算法题目——求一组整数间的总Hamming距离。通过两种不同的实现方式，介绍了如何有效地计算出任意两个数之间的Hamming距离之和，并提供了具体的代码示例。

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前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发优快云，mcf171专栏。

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The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Now your job is to find the total Hamming distance between all pairs of the given numbers.

Example:

Input: 4, 14, 2

Output: 6

Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
showing the four bits relevant in this case). So the answer will be:
HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.

Note:

Elements of the given array are in the range of 0 to 10^9
Length of the array will not exceed 10^4.

这个题目实质上就是找每一位有1的个数数多少，然后0的个数就是数组长度减去1个个数，每一位的组合数就是count_1 * count_0.

解法1：每次都同时计算所有数字的同一位，时间复杂度O(10^4 * 10^9 )，空间复杂度O(10 ^ 4)Your runtime beats 1.26% of java submissions.

public class Solution {
    public int totalHammingDistance(int[] nums) {
        int count = 0;
	List<Integer> data = new ArrayList<Integer>();
	for(int item : nums) data.add(item);
	while(data.size()!=0){
		count +=  diff(data,nums.length);
	}
	return count;
    }

    public int diff(List<Integer> nums, int totalNum){
	int count = 0;
	int count_1 = 0, count_0 = 0;
	for(int i = 0; i < nums.size(); ){
		int val = nums.get(i);
		if(val % 2 == 1) count_1 ++;
		val /= 2;
		if(val == 0)nums.remove(i);
		else{nums.set(i,val); i ++;}
	}
	count_0 = totalNum - count_1;
	count = count_1 * count_0;
	return count;
    }
}

解法2：用一个31位的数组来保存每一位1的数字，时间复杂度O(10^4 * 10^9 )，空间复杂度O(1) Your runtime beats 5.03% of java submissions.

public class Solution {
    public int totalHammingDistance(int[] nums) {
        int[] flags = new int[31];
        for(int item : nums){
            int index = 0;
            while(item != 0){
                if(item %2 != 0) flags[index] ++;
                index ++;
                item = item /2;
            }
        }
        int count = 0;
        for(int item : flags){
            count += item * (nums.length - item);
        }
        return count;
    }
}