[LeetCode] 565. Array Nesting 數組嵌套

時間 2019-11-20

標籤 leetcode array nesting 數組嵌套简体版

原文原文鏈接

A zero-indexed array A of length N contains all integers from 0 to N-1. Find and return the longest length of set S, where S[i] = {A[i], A[A[i]], A[A[A[i]]], ... } subjected to the rule below.html

Suppose the first element in S starts with the selection of element A[i] of index = i, the next element in S should be A[A[i]], and then A[A[A[i]]]… By that analogy, we stop adding right before a duplicate element occurs in S.git

Example 1:github

Input: A = [5,4,0,3,1,6,2]
Output: 4
Explanation: 
A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.

One of the longest S[K]:
S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}

Note:數組

N is an integer within the range [1, 20,000].
The elements of A are all distinct.
Each element of A is an integer within the range [0, N-1].

這道題讓咱們找嵌套數組的最大個數，給的數組總共有n個數字，範圍均在 [0, n-1] 之間，題目中也把嵌套數組的生成解釋的很清楚了，其實就是值變成座標，獲得的數值再變座標。那麼實際上當循環出現的時候，嵌套數組的長度也不能再增長了，而出現的這個相同的數必定是嵌套數組的首元素，博主剛開始沒有想清楚這一點，覺得出現重複數字的地方多是嵌套數組中間的某個位置，因而用個 set 將生成的嵌套數組存入，而後每次查找新生成的數組是否已經存在。並且還以原數組中每一個數字看成嵌套數組的起始數字都算一遍，結果固然是 TLE 了。其實對於遍歷過的數字，咱們不用再將其看成開頭來計算了，而是隻對於未遍歷過的數字看成嵌套數組的開頭數字，不過在進行嵌套運算的時候，並不考慮中間的數字是否已經訪問過，而是隻要找到和起始位置相同的數字位置，而後更新結果 res，參見代碼以下：post

解法一：優化

class Solution {
public:
    int arrayNesting(vector<int>& nums) {
        int n = nums.size(), res = INT_MIN;
        vector<bool> visited(n, false);
        for (int i = 0; i < nums.size(); ++i) {
            if (visited[nums[i]]) continue;
            res = max(res, helper(nums, i, visited));
        }
        return res;
    }
    int helper(vector<int>& nums, int start, vector<bool>& visited) {
        int i = start, cnt = 0;
        while (cnt == 0 || i != start) {
            visited[i] = true;
            i = nums[i];
            ++cnt;
        }
        return cnt;
    }
};

下面這種方法寫法上更簡潔一些，思路徹底同樣，參見代碼以下：url

解法二：spa

class Solution {
public:
    int arrayNesting(vector<int>& nums) {
        int n = nums.size(), res = INT_MIN;
        vector<bool> visited(n, false);
        for (int i = 0; i < n; ++i) {
            if (visited[nums[i]]) continue;
            int cnt = 0, j = i;
            while(cnt == 0 || j != i) {
                visited[j] = true;
                j = nums[j];
                ++cnt;
            }
            res = max(res, cnt);
        }
        return res;
    }
};

下面這種解法是網友 @edyyy 提醒博主的，能夠優化解法二的空間，咱們並不須要專門的數組來記錄數組是否被遍歷過，而是在遍歷的過程當中，將其交換到其應該出現的位置上，由於若是某個數出如今正確的位置上，那麼它必定沒法組成嵌套數組，這樣就至關於咱們標記了其已經訪問過了，思路確實很贊啊，參見代碼以下：code

解法三：htm

class Solution {
public:
    int arrayNesting(vector<int>& nums) {
        int n = nums.size(), res = 0;
        for (int i = 0; i < n; ++i) {
            int cnt = 1;
            while (nums[i] != i) {
                swap(nums[i], nums[nums[i]]);
                ++cnt;
            }
            res = max(res, cnt);
        }
        return res;
    }
};