上岸算法LeetCode Weekly Contest 260解题报告 leetcode算法面试刷题

【 NO.1 增量元素之间的最大差值】
解题思路
遍历数组维护全局最小值，若当前值较大就是一个合理的答案，遍历过程取最大的合理答案即可。
代码展示
public class Solution {

public int maximumDifference(int[] nums) { if (nums == null || nums.length == 0) { return 0; } int res= -1; int minNum = Integer.MAX_VALUE; for (int n : nums) { if (n > minNum) { res = Math.max(n - minNum, res); } minNum = Math.min(minNum, n); } return res; }

}
【 NO.2 网格游戏】
解题思路
注意到网格只有两行，所以第一个机器人需要选择的实际上就是从哪一列向下。在它确定了向下的那一列之后，第二个机器人要么只能拿到第一行开始部分的分数，要么只能拿到第一行结尾部分的分数。
代码展示
public class Solution {

public long gridGame(int[][] grid) { if (grid == null || grid.length == 0 || grid[0].length == 0) { return 0; } int n = grid[0].length; long left = 0, rihgt = 0; for (int i= 1; i < n; i++) { rihgt += grid[0][i]; } long res = rihgt; for (int i = 1; i < n; i++) { left += grid[1][i - 1]; rihgt -= grid[0][i]; res = Math.min(res, Math.max(left, rihgt)); } return res; }

}
【 NO.3 判断单词是否能放入填字游戏内】
解题思路
模拟题，详情见注释。
代码展示
【上岸算法LeetCode Weekly Contest 260解题报告】public class Solution {

public boolean placeWordInCrossword(char[][] board, String word) {if (board == null || board.length == 0 || board[0].length == 0) {return false; }int n = board.length; int m = board[0].length; for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {//从 (i, j) 开始，尝试水平地、垂直地放置单词if (isValid(board, word, j, j + word.length() - 1, i, true) || isValid(board, word, i, i + word.length() - 1, j, false)) {return true; }}}return false; }

private boolean isValid(char[][] board, String word, int start, int end, int standard, boolean isHorizontal) {

//水平放置, standard代表行, 固定不动if (isHorizontal) {if (end > board[0].length - 1) {return false; }//如果左边界不越界，检查左边界的元素是否合法if (start - 1 >= 0 && (board[standard][start - 1] == ' ' || Character.isLetter(board[standard][start - 1]))) {return false; }//如果右边界不越界，检查右边界的元素是否合法if (end + 1 < board[0].length && (board[standard][end + 1] == ' ' || Character.isLetter(board[standard][end + 1]))) {return false; }//至此，它的位置已确认是合法的了//接下来，只需要判断 (standard, start) ~ (standard, end) 这个区间 "是否有障碍'#'//正反都需要判断return check(board, word, start, end, standard, true, false) || check(board, word, start, end, standard, true, true); }//垂直放置,standard 代表列, 固定else {if (end > board.length - 1) {return false; }//如果上边界不越界，检查上边界的元素是否合法if (start - 1 >= 0 && (board[start - 1][standard] == ' ' || Character.isLetter(board[start - 1][standard]))) {return false; }//如果下边界不越界，检查下边界的元素是否合法if (end + 1 < board.length && (board[end + 1][standard] == ' ' || Character.isLetter(board[end + 1][standard]))) {return false; }//至此，它的位置已确认是合法的了//接下来，只需要判断 (start, standard) ~ (end, standard) 这个区间 "是否有障碍'#'//正反都要判断return check(board, word, start, end, standard, false, false) || check(board, word, start, end, standard, false, true); }}

private boolean check(char[][] board, String word, int start, int end, int standard, boolean isHorizontal, boolean isReversed) {

if (isHorizontal) {//正向模拟if (!isReversed) {for (int i = start; i <= end; i++) {if (board[standard][i] == '#' || (Character.isLetter(board[standard][i]) && board[standard][i] != word.charAt(i - start))) {return false; }}}//反向模拟else {for (int i = end; i >= start; i--) {if (board[standard][i] == '#' || (Character.isLetter(board[standard][i]) && board[standard][i] != word.charAt(end - i))) {return false; }}}}else {//正向模拟if (!isReversed) {for (int i = start; i <= end; i++) {if (board[i][standard] == '#' || (Character.isLetter(board[i][standard]) && board[i][standard] != word.charAt(i - start))) {return false; }}}//反向模拟else {for (int i = end; i >= start; i--) {if (board[i][standard] == '#' || (Character.isLetter(board[i][standard]) && board[i][standard] != word.charAt(end - i))) {return false; }}}}return true; }

}
【 NO.4 解出数学表达式的学生分数】
解题思路
首先理一理总体的思路，我们需要做的事情有：

求出表达式的正确值：Stack的方式解决
求出表达式所有可能的值：区间型动态规划解决
计算学生的得分：计分即可
代码展示

public int scoreOfStudents(String s, int[] answers) {

int[] count = new int[1024]; for (int ans : answers) {count[ans]++; }Stack stack = new Stack<>(); stack.push(s.charAt(0) - '0'); for (int i = 1; i < s.length(); i += 2) {// 加法暂时不做，存在栈顶if (s.charAt(i) == '+') {stack.push(s.charAt(i + 1) - '0'); }// 乘法直接运算else {stack.push(stack.pop() * (s.charAt(i + 1) - '0')); }}int rihgtAns = 0; while (stack.size() > 0) {rihgtAns += stack.pop(); }// 计算正确的人数积分int res = count[rihgtAns] * 5; // 枚举所有可能的计算结果int n = s.length(); Set[][] dp = new Set[n + 2][n + 2]; for (int i = 0; i < n + 2; i++) {for (int j = 0; j < n + 2; j++) {dp[i][j] = new HashSet<>(); }}for (int j = 0; j < n; j += 2) {dp[j][j].add(s.charAt(j) - '0'); }// 区间型动态规划for (int len = 2; len < n; len++) {// 左端点 ifor (int i = 0; i + len < n; i += 2) {// 枚举左半部分的长度for (int leftLen = 0; leftLen < len; leftLen += 2) {// left 表示左半部分的值// right 表示右半部分的值for (int left : dp[i][i + leftLen]) {for (int right : dp[i + leftLen + 2][i + len]) {if (s.charAt(i + leftLen + 1) == '+') {if (left + right <= 1000) {dp[i][i + len].add(left + right); }} else {if (left * right <= 1000) {dp[i][i + len].add(left * right); }}}}}}}for (int points : dp[0][n - 1]) {if (points != rihgtAns) {res += 2 * count[points]; }}return res; }