2022.07.18 _ a day

565. 数组嵌套

题目描述

索引从0开始长度为N的数组A,包含0到N - 1的所有整数.找到最大的集合S并返回其大小,其中 S[i] = {A[i], A[A[i]], A[A[A[i]]], ... }且遵守以下的规则.

假设选择索引为i的元素A[i]为S的第一个元素,S的下一个元素应该是A[A[i]],之后是A[A[A[i]]]... 以此类推,不断添加直到S出现重复的元素.

 

示例 1:

输入: A = [5,4,0,3,1,6,2]
输出: 4
解释:
A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.

其中一种最长的 S[K]:
S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}

 

提示：

1. N是[1, 20,000]之间的整数.
2. A中不含有重复的元素.
3. A中的元素大小在[0, N-1]之间.

Related Topics

• 深度优先搜索
• 数组

coding

1. 有向图

n 个不重复数字,范围【0,n - 1】,Can connect one by one or x A ring of directed graph i -> nums[i],Each ring does not exist cross,So once we pass a certain elements,In the future will not be using it again,So I can put it to,You can place to mark it as true.We need to traverse each ring,Get the most ring can be,如果是 isVis = true 的节点, It has been traversal,无需再次遍历,So just considering isVis = false 的节点开始遍历

class Solution {
    
    public int arrayNesting(int[] nums) {
    
        int res = 0;
        boolean[] isVis = new boolean[nums.length];
        for (int i = 0; i < nums.length; i++) {
    
            int cnt = 0;
            while (!isVis[i]) {
    
                isVis[i] = true;
                i = nums[i];
                cnt ++;
            }
            res = Math.max(cnt, res);
        }
        return res;
    }
}

2. In situ tag array

class Solution {
    
    public int arrayNesting(int[] nums) {
    
        int ans = 0, n = nums.length;
        for (int i = 0; i < n; ++i) {
    
            int cnt = 0;
            while (nums[i] < n) {
    
                int num = nums[i];
                nums[i] = n;
                i = num;
                ++cnt;
            }
            ans = Math.max(ans, cnt);
        }
        return ans;
    }
}

3. 深搜 => 超时

class Solution {
    
    public int arrayNesting(int[] nums) {
    
        int max = 0;
        for (int i = 0; i < nums.length; i++) {
    
            if (nums[i] != -1) {
    
                Stack<Integer> stack = new Stack();
                max = Math.max(dfs(nums, i, stack), max);
            }
        }
        return max;
    }
    private int dfs(int[] nums, int index, Stack<Integer> stack) {
    
        if (nums[index] == -1) {
    
            return 0;
        }
        if (stack.contains(nums[index])) {
    
            return stack.size();
        }
        stack.push(nums[index]);
        int size = dfs(nums, nums[index], stack);
        nums[index] = -1;
        stack.pop();
        return size;
    }
}