目录
一、堆
215. 数组中的第K个最大元素
class Solution {
public int findKthLargest(int[] nums, int k) {
PriorityQueue<Integer> maxK=new PriorityQueue<>();
for (int i = 0; i <nums.length ; i++) {
if(i<k){
maxK.add(nums[i]);
}else if(maxK.peek()<nums[i]){
maxK.poll();
maxK.add(nums[i]);
}
}
return maxK.poll();
}
}
347. 前 K 个高频元素
class Solution {
public int[] topKFrequent(int[] nums, int k) {
// 使用字典，统计每个元素出现的次数，元素为键，元素出现的次数为值
HashMap<Integer,Integer> map = new HashMap();
for(int num : nums){
if (map.containsKey(num)) {
map.put(num, map.get(num) + 1);
} else {
map.put(num, 1);
}
}
// 遍历map，用最小堆保存频率最大的k个元素
PriorityQueue<Integer> pq = new PriorityQueue<>(new Comparator<Integer>() {
@Override
public int compare(Integer a, Integer b) {
return map.get(a) - map.get(b);
}
});
for (Integer key : map.keySet()) {
if (pq.size() < k) {
pq.add(key);
} else if (map.get(key) > map.get(pq.peek())) {
pq.remove();
pq.add(key);
}
}
int a[]=new int[k];
for (int i = 0; i <k ; i++) {
a[i]=pq.poll() ;
}
return a;
}
}
二、回溯法
2.1 二维平面回溯
54. 螺旋矩阵
class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
ArrayList<Integer> res=new ArrayList<>();
if(matrix==null||matrix.length==0) return res;
int left=0,right=matrix[0].length-1,up=0,down=matrix.length-1;
while(left<=right&&up<=down){
//从左到右
for(int i=left;i<=right&&up<=down;i++){
res.add(matrix[up][i]);
}
up++;
//从上到下
for(int j=up;j<=down&&left<=right;j++){
res.add(matrix[j][right]);
}
right--;
//从右到左
for(int i=right;i>=left&&up<=down;i--){
res.add(matrix[down][i]);
}
down--;
//从下到上
for(int j=down;j>=up&&left<=right;j--){
res.add(matrix[j][left]);
}
left++;
}
return res;
}
}
剑指 Offer 12. 矩阵中的路径
class Solution {
public boolean exist(char[][] board, String word) {
char[] words = word.toCharArray();
for(int i = 0; i < board.length; i++) {
for(int j = 0; j < board[0].length; j++) {
if(dfs(board, words, i, j, 0)) return true;
}
}
return false;
}
boolean dfs(char[][] board, char[] word, int i, int j, int k) {
if(i >= board.length || i < 0 || j >= board[0].length || j < 0 || board[i][j] != word[k]) return false;
if(k == word.length - 1) return true;
char tmp = board[i][j];
board[i][j] = '/';
boolean res = dfs(board, word, i + 1, j, k + 1) || dfs(board, word, i - 1, j, k + 1) ||
dfs(board, word, i, j + 1, k + 1) || dfs(board, word, i , j - 1, k + 1);
board[i][j] = tmp;
return res;
}
}
剑指 Offer 13. 机器人的运动范围
class Solution {
//原来是求能走到的所有格子之和.简单的BFS,每加一个合法的坐标, res += 1.
int count=0;
public int movingCount(int m, int n, int k) {
boolean[][] visited=new boolean[m][n];
return recur(0,0,m,n,k,visited);
}
public int recur(int i,int j,int m,int n,int k, boolean visited[][]){
//当在方格范围外面或者已经访问过或者方格之和大于18,则为0
if(i<0||i>=m||j<0||j>=n||(i/10+i%10+j/10+j%10)>k||visited[i][j]==true){
return 0;
}
count+=1;
visited[i][j] = true;
recur(i+1,j,m,n,k,visited);
recur(i-1,j,m,n,k,visited);
recur(i,j-1,m,n,k,visited);
recur(i,j+1,m,n,k,visited);
return count;
}
}
200. 岛屿数量
class Solution {
int count=0;
public int numIslands(char[][] grid) {
for(int i=0;i<grid.length;i++){
for(int j=0;j<grid[0].length;j++){
if(grid[i][j]=='1'){
count++;
infect(grid,i,j);
}
}
}
return count;
}
public void infect(char[][] grid,int i,int j){
if(i<0||j<0||i>=grid.length||j>=grid[0].length||grid[i][j]!='1') return;
grid[i][j]='2';
infect(grid,i+1,j);
infect(grid,i,j+1);
infect(grid,i-1,j);
infect(grid,i,j-1);
}
}
2.2 排列组合
78. 子集
class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> res=new ArrayList<>();
res.add(new ArrayList<>());
for(int i=0;i<nums.length;i++){
int n=res.size();
for(int j=0;j<n;j++){
List<Integer> list=new ArrayList<>(res.get(j));
if (i >0 && nums[i] == nums[i - 1]) {
continue;
}
list.add(nums[i]);
res.add(list);
}
}
return res;
}
}
46. 全排列
77. 组合
给定两个整数 n 和 k，返回 1 … n 中所有可能的 k 个数的组合。
