Java算法练习5-编程知识

Java算法练习5

- 1.8 [268. 丢失的数字](https://leetcode.cn/problems/missing-number/)
- 1.9 [383. 赎金信](https://leetcode.cn/problems/ransom-note/)
- 1.10 [2133. 检查是否每一行每一列都包含全部整数](https://leetcode.cn/problems/check-if-every-row-and-column-contains-all-numbers/)
- 1.11 [599. 两个列表的最小索引总和](https://leetcode.cn/problems/minimum-index-sum-of-two-lists/)
- 1.12 [36. 有效的数独](https://leetcode.cn/problems/valid-sudoku/)
- 1.13 [17. 电话号码的字母组合](https://leetcode.cn/problems/letter-combinations-of-a-phone-number/)
- 1.14 [37. 解数独](https://leetcode.cn/problems/sudoku-solver/)

1.8 268. 丢失的数字

给定一个包含 [0, n] 中 n 个数的数组 nums ，找出 [0, n] 这个范围内没有出现在数组中的那个数。

示例 1：

输入：nums = [3,0,1]
输出：2
解释：n = 3，因为有 3 个数字，所以所有的数字都在范围 [0,3] 内。2 是丢失的数字，因为它没有出现在 nums 中。

示例 2：

输入：nums = [0,1]
输出：2
解释：n = 2，因为有 2 个数字，所以所有的数字都在范围 [0,2] 内。2 是丢失的数字，因为它没有出现在 nums 中。

示例 3：

输入：nums = [9,6,4,2,3,5,7,0,1]
输出：8
解释：n = 9，因为有 9 个数字，所以所有的数字都在范围 [0,9] 内。8 是丢失的数字，因为它没有出现在 nums 中。

示例 4：

输入：nums = [0]
输出：1
解释：n = 1，因为有 1 个数字，所以所有的数字都在范围 [0,1] 内。1 是丢失的数字，因为它没有出现在 nums 中。

class Solution {public int missingNumber(int[] nums) {int n = nums.length;Set<Integer> set = new HashSet<Integer>();for(int i = 0;i< n;i++){set.add(nums[i]);}int missNum = -1;for(int j = 0;j <= n; j++){if(!set.contains(j)){missNum = j;break;}}return missNum;}
}

1.9 383. 赎金信

给你两个字符串：ransomNote 和 magazine ，判断 ransomNote 能不能由 magazine 里面的字符构成。

如果可以，返回 true ；否则返回 false 。

magazine 中的每个字符只能在 ransomNote 中使用一次。

示例 1：
输入：ransomNote = "a", magazine = "b"
输出：false
示例 2：
输入：ransomNote = "aa", magazine = "ab"
输出：false
示例 3：
输入：ransomNote = "aa", magazine = "aab"
输出：true

class Solution {public boolean canConstruct(String ransomNote, String magazine) {int[] s = new int[26];for(char c : magazine.toCharArray()){s[c-'a']++;}for(char c : ransomNote.toCharArray()){s[c-'a']--;if(s[c-'a'] < 0){return false;}}return true;}
}

1.10 2133. 检查是否每一行每一列都包含全部整数

对一个大小为 n x n 的矩阵而言，如果其每一行和每一列都包含从 1 到 n 的全部整数（含 1 和 n），则认为该矩阵是一个有效矩阵。

给你一个大小为 n x n 的整数矩阵 matrix ，请你判断矩阵是否为一个有效矩阵：如果是，返回 true ；否则，返回 false 。

示例 1：
输入：matrix = [[1,2,3],[3,1,2],[2,3,1]]
输出：true
解释：在此例中，n = 3 ，每一行和每一列都包含数字 1、2、3 。
因此，返回 true 。
示例 2：
输入：matrix = [[1,1,1],[1,2,3],[1,2,3]]
输出：false
解释：在此例中，n = 3 ，但第一行和第一列不包含数字 2 和 3 。
因此，返回 false 。

class Solution {public boolean checkValid(int[][] matrix) {int n = matrix.length;for(int i = 0;i < n ; i++){Set<Integer> s1 = new HashSet();Set<Integer> s2 = new HashSet();for(int j = 0;j < n ; j++){s1.add( matrix[j][i]);s2.add( matrix[i][j]);}if(s1.size() != n || s2.size() != n){return false;}}return true;}
}

1.11 599. 两个列表的最小索引总和

假设 Andy 和 Doris 想在晚餐时选择一家餐厅，并且他们都有一个表示最喜爱餐厅的列表，每个餐厅的名字用字符串表示。

你需要帮助他们用最少的索引和找出他们共同喜爱的餐厅。如果答案不止一个，则输出所有答案并且不考虑顺序。你可以假设答案总是存在。

示例 1:
输入: list1 = ["Shogun", "Tapioca Express", "Burger King", "KFC"]，list2 = ["Piatti", "The Grill at Torrey Pines", "Hungry Hunter Steakhouse", "Shogun"]
输出: ["Shogun"]
解释: 他们唯一共同喜爱的餐厅是“Shogun”。
示例 2:
输入:list1 = ["Shogun", "Tapioca Express", "Burger King", "KFC"]，list2 = ["KFC", "Shogun", "Burger King"]
输出: ["Shogun"]
解释: 他们共同喜爱且具有最小索引和的餐厅是“Shogun”，它有最小的索引和1(0+1)。

class Solution {public String[] findRestaurant(String[] list1, String[] list2) {Map<String,Integer> map = new HashMap();for(int i = 0;i < list1.length;i++){map.put(list1[i],i);}List<String> list = new ArrayList();int minIndex = 2000;for(int i = 0;i < list2.length;i++){if(map.containsKey(list2[i])){int index = map.get(list2[i]) + i;if(index < minIndex){list.clear();minIndex = index;list.add(list2[i]);}else if(index == minIndex){list.add(list2[i]);}}}return list.toArray(new String[list.size()]);}
}

1.12 36. 有效的数独

请你判断一个 9 x 9 的数独是否有效。只需要 根据以下规则 ，验证已经填入的数字是否有效即可。

数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图）

注意：

一个有效的数独（部分已被填充）不一定是可解的。
只需要根据以上规则，验证已经填入的数字是否有效即可。
空白格用 '.' 表示。

示例 1：
输入：board = 
[["5","3",".",".","7",".",".",".","."]
,["6",".",".","1","9","5",".",".","."]
,[".","9","8",".",".",".",".","6","."]
,["8",".",".",".","6",".",".",".","3"]
,["4",".",".","8",".","3",".",".","1"]
,["7",".",".",".","2",".",".",".","6"]
,[".","6",".",".",".",".","2","8","."]
,[".",".",".","4","1","9",".",".","5"]
,[".",".",".",".","8",".",".","7","9"]]
输出：true
示例 2：
输入：board = 
[["8","3",".",".","7",".",".",".","."]
,["6",".",".","1","9","5",".",".","."]
,[".","9","8",".",".",".",".","6","."]
,["8",".",".",".","6",".",".",".","3"]
,["4",".",".","8",".","3",".",".","1"]
,["7",".",".",".","2",".",".",".","6"]
,[".","6",".",".",".",".","2","8","."]
,[".",".",".","4","1","9",".",".","5"]
,[".",".",".",".","8",".",".","7","9"]]
输出：false
解释：除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

class Solution {public boolean isValidSudoku(char[][] board) {int[][] rows = new int[9][9];int[][] columns = new int[9][9];int[][][] subboxes = new int[3][3][9];for(int i = 0;i < 9 ;i++){for(int j = 0;j < 9 ;j++){char c = board[i][j];if(c!='.'){int index = c - '1';rows[i][index]++;columns[j][index]++;subboxes[i / 3][j / 3 ][index]++;if(rows[i][index] >1 || columns[j][index] >1 ||subboxes[i / 3][j / 3 ][index]>1){return false;}}}}return true;}
}

1.13 17. 电话号码的字母组合

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 任意顺序 返回。

给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

示例 1：
输入：digits = "23"
输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]
示例 2：
输入：digits = ""
输出：[]
示例 3：
输入：digits = "2"
输出：["a","b","c"]

class Solution {String[] letterMap = {" "," ","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};List<String> list = new ArrayList();public List<String> letterCombinations(String digits) {if(digits.equals(""))return list;findCombination(digits, 0, new StringBuilder());return list;}private void findCombination(String digits, int index, StringBuilder s){if(index == digits.length()){list.add(s.toString());return;}char c = digits.charAt(index);String letters = letterMap[c - '0'];for(int i = 0 ; i < letters.length() ; i++){findCombination(digits, index+1, s.append(letters.charAt(i)));s.deleteCharAt(s.length()-1);}}}

1.14 37. 解数独

编写一个程序，通过填充空格来解决数独问题。

数独的解法需 遵循如下规则：

数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图）

数独部分空格内已填入了数字，空白格用 '.' 表示。

示例 1：
输入：board = [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"]]
输出：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：

class Solution {public void solveSudoku(char[][] board) {solve(board);}public boolean solve(char[][] board){for(int i = 0;i < 9;i++){for(int j = 0;j < 9;j++){if(board[i][j] != '.'){continue;}for(char k ='1' ;k <= '9';k++){if(isVailSudoku(i,j,k,board)){board[i][j] = k;if(solve(board)){return true;}else{board[i][j] = '.';}}}return false;}}return true;}public boolean isVailSudoku(int row, int col,char val,char[][] board){for(int i = 0;i < 9;i++){if(board[row][i] == val) return false;}for(int i = 0;i < 9;i++){if(board[i][col] == val) return false;}int startRow = (row / 3) * 3; int startCol = (col / 3) * 3;for (int i = startRow; i < startRow + 3; i++){for (int j = startCol; j < startCol + 3; j++){if (board[i][j] == val){return false;}}}return true;}
}