No.1
判断字符串的两半是否类似
解题思路
统计元音字母数量即可。
代码展现
class Solution { public boolean halvesAreAlike(String s) { int n = s.length(); int a = 0, b = 0; for (int i = 0, j = n - 1; i < j; i++, j--) { a += "AEIOUaeiou".indexOf(s.charAt(i)) >= 0 ? 1 : 0; b += "AEIOUaeiou".indexOf(s.charAt(j)) >= 0 ? 1 : 0; } return a == b; }}No.2
吃苹果的最大数目
解题思路
贪婪的思路，咱们总是吃掉剩下的苹果中最先烂掉的。应用优先队列能够保护剩下的苹果哪些先烂掉。
代码展现
class Solution { static class Apple { int num; int day; public Apple(int num, int day) { this.num = num; this.day = day; } } public int eatenApples(int[] apples, int[] days) { PriorityQueue<Apple> bucket = new PriorityQueue<>(Comparator.comparingInt(a -> a.day)); int res = 0; for (int i = 0; i < apples.length; i++) { if (apples[i] > 0) { bucket.add(new Apple(apples[i], i + days[i])); } res += eat(bucket, i); } // 不再减少苹果，将剩下的吃完 for (int i = apples.length; !bucket.isEmpty(); i++) { res += eat(bucket, i); } return res; } // 在第 day 天吃苹果，返回能吃到的数量 (0 或 1) private int eat(PriorityQueue<Apple> bucket, int day) { while (!bucket.isEmpty()) { Apple a = bucket.poll(); if (a.day <= day) { continue; } if ((--a.num) > 0) { bucket.add(a); } return 1; } return 0; }}No.3
球会落何处
解题思路
模仿球的着落即可。
代码展现
class Solution { public int[] findBall(int[][] grid) { int m = grid[0].length; int[] res = new int[m]; for (int i = 0; i < m; i++) { res[i] = drop(0, i, grid); } return res; } private int drop(int x, int y, int[][] grid) { if (x == grid.length) { return y; } if (grid[x][y] == 1 && (y == grid[0].length - 1 || grid[x][y + 1] == -1)) { return -1; } if (grid[x][y] == -1 && (y == 0 || grid[x][y - 1] == 1)) { return -1; } return drop(x + 1, y + grid[x][y], grid); }}No.4
与数组中元素的最大异或值
解题思路
字典树。
咱们晓得，二进制数的异或运算的含意就是两者是否不同 —— 所以咱们在 Trie 树上尽可能走与以后位不同的那一条门路即可。
代码展现
class Solution { class TrieNode { int min; // 以后节点下最小的数 TrieNode[] child; public TrieNode() { min = Integer.MAX_VALUE; child = new TrieNode[2]; } } public int[] maximizeXor(int[] nums, int[][] queries) { int min = Arrays.stream(nums).min().getAsInt(); // 初始化，建设一棵 32 层的 Trie 树 TrieNode root = new TrieNode(); for (int num : nums) { TrieNode cur = root; for (int m = 30; m >= 0; m--) { cur.min = Math.min(cur.min, num); int d = (num >> m) & 1; if (cur.child[d] == null) { cur.child[d] = new TrieNode(); } cur = cur.child[d]; cur.min = Math.min(cur.min, num); } } // 求解 int[] res = new int[queries.length]; for (int i = 0; i < queries.length; i++) { if (queries[i][1] < min) { res[i] = -1; continue; } TrieNode cur = root; for (int m = 30; m >= 0 && cur != null; m--) { int xd = (queries[i][0] >> m) & 1; // 心愿往 xd ^ 1 的方向走 TrieNode except = cur.child[xd ^ 1]; if (except == null || except.min > queries[i][1]) { cur = cur.child[xd]; } else { cur = except; } } res[i] = cur == null || cur.min > queries[i][1] ? -1 : cur.min ^ queries[i][0]; } return res; }}