Largest Divisible Subset

https://leetcode.com/problems/largest-divisible-subset/description/

Given a set of distinct positive integers, find the largest subset such that every pair (Si, Sj) of elements in this subset satisfies: Si % Sj = 0 or Sj % Si = 0.
If there are multiple solutions, return any subset is fine.

Thoughts

如果这道题只是问最长的能互相除的子序列是什么，会简洁很多，思路和LIS相似。让f[i]表示以i为尾的互除子序列，f[i]初始值为1，因为它能除自身。然后遍历所有已经遍历过了的看nums[i]是否能整除nums[j], 如果可以整除则把f[j] + 1和现有f[i]比较取最大值。这题比较麻烦的是不但让你返回最大长度，还要求返回完整的序列。因此我们用一个array记录下对于f[i]找到的最大的j。然后就能从尾往前回溯得到目标序列了。

Code

class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        int n = nums.length, max = 0, maxIndex = 0;
        int[] f = new int[n], pres = new int[n];
        List<Integer> res = new ArrayList<>();
        if (n == 0) {
            return res;
        }
        Arrays.sort(nums);
        for (int i = 0; i < nums.length; i++) {
            f[i] = 1;
            pres[i] = i;
            for (int j = 0; j < i; j++) {
                if (nums[i] % nums[j] == 0 && f[j] + 1 > f[i]) {
                    pres[i] = j; 
                    f[i] = f[j] + 1;
                }
            }
            if (f[i] > max) {
                max = f[i];
                maxIndex = i;
            }
        }

        while (pres[maxIndex] != maxIndex) {
            res.add(nums[maxIndex]);
            maxIndex = pres[maxIndex];
        }
        res.add(nums[maxIndex]);
        return res;
    }
}

Analysis

Errors:

max = f[i];忘了写
res.add(nums[subres])写在了map.get后
没有排序

做题耗时50min

时间复杂度O(n^2).

PreviousWiggle Subsequence NextMaximum Length of Pair Chain

Last updated 5 years ago

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Thoughts

Code

class Solution {
    public List<Integer> largestDivisibleSubset(int[] nums) {
        int n = nums.length, max = 0, maxIndex = 0;
        int[] f = new int[n], pres = new int[n];
        List<Integer> res = new ArrayList<>();
        if (n == 0) {
            return res;
        }
        Arrays.sort(nums);
        for (int i = 0; i < nums.length; i++) {
            f[i] = 1;
            pres[i] = i;
            for (int j = 0; j < i; j++) {
                if (nums[i] % nums[j] == 0 && f[j] + 1 > f[i]) {
                    pres[i] = j; 
                    f[i] = f[j] + 1;
                }
            }
            if (f[i] > max) {
                max = f[i];
                maxIndex = i;
            }
        }

        while (pres[maxIndex] != maxIndex) {
            res.add(nums[maxIndex]);
            maxIndex = pres[maxIndex];
        }
        res.add(nums[maxIndex]);
        return res;
    }
}