技巧题
只出现一次的数字
class Solution {/** HashMap统计num和次数前面一天用堆魔怔了，也用堆进行排列...*/public int singleNumber(int[] nums) {PriorityQueue<int[]> pq = new PriorityQueue<>(new Comparator<int[]>() {public int compare(int[] o1, int[] o2) {return o1[1] - o2[1];}});HashMap<Integer, Integer> map = new HashMap<>();for(int num : nums) {map.put(num, map.getOrDefault(num, 0) + 1);}for(var node : map.entrySet()) {int[] temp = new int[2];temp[0] = node.getKey();temp[1] = node.getValue();pq.offer(temp);}return pq.peek()[0];}
}
多数元素
class Solution {/** HashMap统计?*/public int majorityElement(int[] nums) {int res = 0;int n = nums.length / 2;HashMap<Integer, Integer> map = new HashMap<>();for(int num: nums) {map.put(num, map.getOrDefault(num, 0) + 1);}for(var node: map.entrySet()) {if(node.getValue() > n) {res = node.getKey();}}return res;}
}
颜色分类
class Solution {/** 使用堆不就秒了?*/public void sortColors(int[] nums) {PriorityQueue<Integer> pq = new PriorityQueue<>();for(int num : nums) {pq.offer(num);}int i = 0;while(!pq.isEmpty()) {nums[i] = pq.poll();i++;}}
}
芜湖~ 力扣200题打卡！
下一个排列
class Solution {/** i = 0 1 2 32 1 4 3 从后向前搜索2|1 3 4 搜到第一个i > i - 1的，记录i，排序i到末尾2|3 1 4 找到排序后第一个 > i - 1的，交换需要注意Arrays.sort(nums, i, j) 是左闭右开的区间对nums进行从i到末尾的sort需要(nums, i, len(而不是len - 1))*/public void nextPermutation(int[] nums) {int len = nums.length;for(int i = len - 1; i >= 0; i--) {if( i > 0 && nums[i] > nums[i - 1]) {Arrays.sort(nums, i, len);for(int j = i; j < len; j++) {if(nums[j] > nums[i - 1]) {int temp = nums[i - 1];nums[i - 1] = nums[j];nums[j] = temp;return;}}}else if(i == 0) {Arrays.sort(nums);} }}
}
寻找重复数
class Solution {/** hashmap直接秒了啊*/public int findDuplicate(int[] nums) {int res = 0;HashMap<Integer, Integer> map = new HashMap<>();for(int num: nums) {map.put(num, map.getOrDefault(num, 0) + 1);}for(var node: map.entrySet()) {if(node.getValue() > 1) {res = node.getKey();}}return res;}
}
图论
岛屿数量
class Solution {/** 深度优先搜索dfs类似和寻找字符串逐行逐列遍历判断*/public int numIslands(char[][] grid) {int row = grid.length;int col = grid[0].length;int island = 0;for(int i = 0; i < row; i++) {for(int j = 0; j < col; j++) {if(grid[i][j] == '1') {island++;dfs(grid, i, j);}}}return island;}public void dfs(char[][] grid, int i, int j) {int row = grid.length;int col = grid[0].length;if(i >= row || i < 0 || j >= col || j < 0 || grid[i][j] == '0') {return;}grid[i][j] = '0';dfs(grid, i + 1, j);dfs(grid, i, j + 1);dfs(grid, i - 1, j);dfs(grid, i, j - 1);}
}
腐烂的橘子
class Solution {/** 用2覆盖grid网格2表示初始化时间*/public int orangesRotting(int[][] grid) {if(grid == null || grid.length == 0) {return 0;}int row = grid.length;int col = grid[0].length;for(int i = 0; i < row; i++) {for(int j = 0; j < col; j++) {if(grid[i][j] == 2) {dfs(grid, i, j, 2);}}}int maxtime = 0;for(int i = 0; i < row; i++) {for(int j = 0; j < col; j++) {if(grid[i][j] == 1) {return -1;}else {maxtime = Math.max(maxtime, grid[i][j]);}}}return maxtime == 0 ? 0 : maxtime - 2;}public void dfs(int[][] grid, int i, int j, int time) {int row = grid.length;int col = grid[0].length;if(i >= row || i < 0 || j >= col || j < 0) {return;}if(grid[i][j] != 1 && grid[i][j] < time) { // 没新鲜橘子，且时间更少，倒回去return;}grid[i][j] = time;dfs(grid, i + 1, j, time + 1);dfs(grid, i, j + 1, time + 1);dfs(grid, i - 1, j, time + 1);dfs(grid, i, j - 1, time + 1);}
}
课程表
实现Trie 前缀树
class Trie {Map<Character, Map> map, t;public Trie() {map = new HashMap<>();}public void insert(String word) {t = map;for(char w: word.toCharArray()) {if(!t.containsKey(w)) {t.put(w, new HashMap<Character, Map>());}t = t.get(w);}t.put('#', null);}public boolean search(String word) {t = map;for(char w: word.toCharArray()) {if(!t.containsKey(w)) {return false;}t = t.get(w);}return t.containsKey('#');}public boolean startsWith(String prefix) {t = map;for(char w: prefix.toCharArray()) {if(!t.containsKey(w)) {return false;}t = t.get(w);}return true;}
}/*** Your Trie object will be instantiated and called as such:* Trie obj = new Trie();* obj.insert(word);* boolean param_2 = obj.search(word);* boolean param_3 = obj.startsWith(prefix);*/
)
