力扣热题 100
哈希
双指针
滑动窗口
子串
普通数组
矩阵
链表
二叉树
- 二叉树的中序遍历
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> reusltList = new ArrayList<>();
inorderVisited(root, reusltList);
return reusltList;
}
private void inorderVisited(TreeNode root, List<Integer> reusltList) {
if (root == null) {
return;
}
inorderVisited(root.left, reusltList);
reusltList.add(root.val);
inorderVisited(root.right, reusltList);
}
}
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> reusltList = new ArrayList<>();
inorderVisited(root, reusltList);
return reusltList;
}
private void inorderVisited(TreeNode root, List<Integer> reusltList) {
Stack<TreeNode> stack = new Stack<>();
while (root != null || !stack.empty()) {
while (root != null) {
stack.push(root);
root = root.left;
}
root = stack.pop();
reusltList.add(root.val);
root = root.right;
}
}
}
图论
回溯
二分查找
栈
堆
贪心算法
动态规划
- 爬楼梯
class Solution {
public int climbStairs(int n) {
int[] dp = new int[n+1];
dp[0] = 1;
dp[1] = 1;
for(int i = 2;i <= n; i++){
dp[i] = dp[i-1] + dp[i-2];
}
return dp[n];
}
}
- 使用最小花费爬楼梯
class Solution {
public int minCostClimbingStairs(int[] cost) {
int len = cost.length;
int[] dp = new int[len + 1];
dp[0] = 0;
dp[1] = 0;
for (int i = 2; i <= len; i++) {
dp[i] = Math.min((dp[i - 1] + cost[i - 1]), (dp[i - 2] + cost[i - 2]));
}
return dp[len];
}
}
- 打家劫舍
class Solution {
public int rob(int[] nums) {
if (nums.length == 0) {
return 0;
}
int len = nums.length;
int[] dp = new int[len+1];
dp[0] = 0;
dp[1] = nums[0];
for(int i = 2; i <= len; i++){
dp[i] = Math.max(dp[i-1], dp[i-2] + nums[i-1]);
}
return dp[len];
}
}
- 完全平方数
class Solution {
public int numSquares(int n) {
int[] dp = new int[n + 1];
for(int i = 1; i <= n;i++) {
int minn = Integer.MAX_VALUE;
for(int j = 1; j * j <= i; j++){
minn = Math.min(minn, dp[i - j * j]);
}
dp[i] = minn + 1;
}
return dp[n];
}
}
- 零钱兑换
class Solution {
public int coinChange(int[] coins, int amount) {
int[] dp = new int[amount + 1];
Arrays.fill(dp, amount + 1);
dp[0] = 0;
for (int i = 1; i <= amount; i++) {
for (int coin : coins) {
if (i < coin) {
continue;
}
dp[i] = Math.min(dp[i], dp[i - coin]);
}
dp[i] += 1;
}
return dp[amount] > amount ? -1 : dp[amount];
}
}
class Solution {
public int coinChange(int[] coins, int amount) {
int[] dp = new int[amount + 1];
Arrays.fill(dp, amount + 1);
dp[0] = 0;
for (int i = 1; i <= amount; i++) {
for (int coin : coins) {
if (i < coin) {
continue;
}
dp[i] = Math.min(dp[i], dp[i - coin] + 1);
}
}
return (dp[amount] == amount + 1) ? -1 : dp[amount];
}
}
01背包
public class BagProblem {
public static void main(String[] args) {
int[] weight = {1,3,4};
int[] value = {15,20,30};
int bagSize = 4;
testWeightBagProblem(weight,value,bagSize);
}
/**
* 动态规划获得结果
* @param weight 物品的重量
* @param value 物品的价值
* @param bagSize 背包的容量
*/
public static void testWeightBagProblem(int[] weight, int[] value, int bagSize){
// 创建dp数组
int goods = weight.length; // 获取物品的数量
int[][] dp = new int[goods][bagSize + 1];
// 初始化dp数组
// 创建数组后，其中默认的值就是0
for (int j = weight[0]; j <= bagSize; j++) {
dp[0][j] = value[0];
}
// 填充dp数组
for (int i = 1; i < weight.length; i++) {
for (int j = 1; j <= bagSize; j++) {
if (j < weight[i]) {
/**
* 当前背包的容量都没有当前物品i大的时候，是不放物品i的
* 那么前i-1个物品能放下的最大价值就是当前情况的最大价值
*/
dp[i][j] = dp[i-1][j];
} else {
/**
* 当前背包的容量可以放下物品i
* 那么此时分两种情况：
* 1、不放物品i
* 2、放物品i
* 比较这两种情况下，哪种背包中物品的最大价值最大
*/
dp[i][j] = Math.max(dp[i-1][j] , dp[i-1][j-weight[i]] + value[i]);
}
}
}
// 打印dp数组
for (int i = 0; i < goods; i++) {
for (int j = 0; j <= bagSize; j++) {
System.out.print(dp[i][j] + "\t");
}
System.out.println("\n");
}
}
}
01背包-滚动数组
public static void main(String[] args) {
int[] weight = {1, 3, 4};
int[] value = {15, 20, 30};
int bagWight = 4;
testWeightBagProblem(weight, value, bagWight);
}
public static void testWeightBagProblem(int[] weight, int[] value, int bagWeight){
int wLen = weight.length;
//定义dp数组：dp[j]表示背包容量为j时，能获得的最大价值
int[] dp = new int[bagWeight + 1];
//遍历顺序：先遍历物品，再遍历背包容量
for (int i = 0; i < wLen; i++){
for (int j = bagWeight; j >= weight[i]; j--){
dp[j] = Math.max(dp[j], dp[j - weight[i]] + value[i]);
}
}
//打印dp数组
for (int j = 0; j <= bagWeight; j++){
System.out.print(dp[j] + " ");
}
}
- 分割等和子集
class Solution {
public boolean canPartition(int[] nums) {
if (nums == null || nums.length == 0) {
return false;
}
int len = nums.length;
int sum = 0;
for (int num : nums) {
sum += num;
}
if (sum % 2 != 0)
return false;
int target = sum / 2;
int[] dp = new int[target + 1];
for (int i = 0; i < len; i++) {
for (int j = target; j >= nums[i]; j--) {
dp[j] = Math.max(dp[j], dp[j - nums[i]] + nums[i]);
}
if (dp[target] == target) {
return true;
}
}
return dp[target] == target;
}
}
- 最后一块石头的重量 II
class Solution {
public int lastStoneWeightII(int[] stones) {
int len = stones.length;
int sum = 0;
for (int stone : stones) {
sum += stone;
}
int target = sum / 2;
int[] dp = new int[target + 1];
for (int i = 0; i < len; i++) {
for (int j = target; j >= stones[i]; j--) {
dp[j] = Math.max(dp[j], dp[j - stones[i]] + stones[i]);
}
}
return sum - dp[target] - dp[target];
}
}