You are given an array of distinct integers arr and an array of integer arrays pieces, where the integers in pieces are distinct. Your goal is to form arr by concatenating the arrays in pieces in any order. However, you are not allowed to reorder the integers in each array pieces[i].

Return true if it is possible to form the array arr from pieces. Otherwise, return false.

Example 1:

Input: arr = [85], pieces = [[85]]
Output: true

Example 2:

Input: arr = [15,88], pieces = [[88],[15]]
Output: true
Explanation: Concatenate [15] then [88]

Example 3:

Input: arr = [49,18,16], pieces = [[16,18,49]]
Output: false
Explanation: Even though the numbers match, we cannot reorder pieces[0].

Example 4:

Input: arr = [91,4,64,78], pieces = [[78],[4,64],[91]]
Output: true
Explanation: Concatenate [91] then [4,64] then [78]

Example 5:

Input: arr = [1,3,5,7], pieces = [[2,4,6,8]]
Output: false 

Constraints:

• 1 <= pieces.length <= arr.length <= 100
• sum(pieces[i].length) == arr.length
• 1 <= pieces[i].length <= arr.length
• 1 <= arr[i], pieces[i][j] <= 100
• The integers in arr are distinct.
• The integers in pieces are distinct (i.e., If we flatten pieces in a 1D array, all the integers in this array are distinct).

能否连接形成数组。

题目说的很绕，给的input中，arr 是一个一维数组，pieces 是一个二维数组，请问你在不破坏 pieces 里面任何一个子数组元素之间相对顺序的情况下，能否将pieces重新排列，使得重新排列之后的 pieces 里面的元素的顺序能和 arr 一样。

这道题不涉及任何算法，基本思路就是判断 pieces 里面是否有任何一个子数组之间元素的相对顺序不符合 arr 里面这些元素的相对顺序。我这里给出两种实现，一种利用到StringBuilder，一种利用到hashmap。时间空间复杂度均为O(n)。

StringBuilder

Java实现

 1 class Solution {
 2     public boolean canFormArray(int[] arr, int[][] pieces) {
 3         StringBuilder sb = new StringBuilder();
 4         for (int num : arr) {
 5             sb.append("#");
 6             sb.append(num);
 7             sb.append("#");
 8         }
 9 
10         for (int i = 0; i < pieces.length; i++) {
11             StringBuilder res = new StringBuilder();
12             for (int j = 0; j < pieces[i].length; j++) {
13                 res.append("#");
14                 res.append(String.valueOf(pieces[i][j]));
15                 res.append("#");
16             }
17             if (!sb.toString().contains(res.toString())) {
18                 return false;
19             }
20         }
21         return true;
22     }
23 }

 

hashmap

Java实现

 1 class Solution {
 2     public boolean canFormArray(int[] arr, int[][] pieces) {
 3         HashMap<Integer, int[]> map = new HashMap<>();
 4         for (int[] piece : pieces) {
 5             map.put(piece[0], piece);
 6         }
 7 
 8         int i = 0;
 9         while (i < arr.length) {
10             if (map.containsKey(arr[i])) {
11                 int[] piece = map.get(arr[i]);
12                 for (int j = 0; j < piece.length; j++) {
13                     if (arr[i] != piece[j]) {
14                         return false;
15                     } else {
16                         i++;
17                     }
18                 }
19             } else {
20                 return false;
21             }
22         }
23         return true;
24     }
25 }

 

LeetCode 题目总结