从今天开始用Java写了
Easy
[338] 比特位计数
dynamic-programming | bit-manipulation
2022/04/07
class Solution {
public int[] countBits(int n) {
int k = 1;
int[] ans = new int[n + 1];
for (int i = 1; i <= n; i++) {
if (i == k) {
ans[i] = 1;
k *= 2;
} else {
ans[i] = 1 + ans[i - k / 2];
}
}
return ans;
}
}
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Medium
[279] 完全平方数
math | dynamic-programming | breadth-first-search
2022/04/05
class Solution {
public int numSquares(int n) {
int[] dp = new int[n];
int k = 1;
for (int i = 0; i < n; i++) {
if (i + 1 == k * k) {
dp[i] = 1;
k++;
} else {
dp[i] = Integer.MAX_VALUE;
for (int j = 1; j < k; j++) {
dp[i] = Math.min(dp[i], 1 + dp[i - j * j]);
}
}
}
return dp[n - 1];
}
}
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[287] 寻找重复数
array | two-pointers | binary-search
2022/04/05
class Solution {
public int findDuplicate(int[] nums) {
int fast = nums[nums[0]], slow = nums[0];
while (fast != slow) {
fast = nums[nums[fast]];
slow = nums[slow];
}
fast = 0;
while (nums[slow] != nums[fast]) {
fast = nums[fast];
slow = nums[slow];
}
return nums[slow];
}
}
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[309] 最佳买卖股票时机含冷冻期
dynamic-programming
2022/04/06
class Solution {
public int maxProfit(int[] prices) {
int n = prices.length;
int[] buy = new int[n];
int[] sell = new int[n];
int[] frozen = new int[n];
buy[0] -= prices[0];
for (int i = 1; i < n; i++) {
buy[i] = Math.max(frozen[i - 1] - prices[i], buy[i - 1]);
sell[i] = Math.max(buy[i - 1] + prices[i], sell[i - 1]);
frozen[i] = Math.max(sell[i - 1], frozen[i - 1]);
}
return Math.max(buy[n - 1], Math.max(sell[n - 1], frozen[n - 1]));
}
}
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[337] 打家劫舍 III
tree | depth-first-search
2022/04/06
class Solution {
private Map<TreeNode, Integer> chosenDp = new HashMap<>();
private Map<TreeNode, Integer> noChosenDp = new HashMap<>();
public int rob(TreeNode root) {
dfs(root);
return Math.max(chosenDp.getOrDefault(root, 0), noChosenDp.getOrDefault(root, 0));
}
private void dfs(TreeNode node) {
if (node == null) {
return;
} else {
dfs(node.left);
dfs(node.right);
chosenDp.put(node,
node.val + noChosenDp.getOrDefault(node.left, 0) + noChosenDp.getOrDefault(node.right, 0));
noChosenDp.put(node, Math.max(chosenDp.getOrDefault(node.left, 0), noChosenDp.getOrDefault(node.left, 0))
+ Math.max(chosenDp.getOrDefault(node.right, 0), noChosenDp.getOrDefault(node.right, 0)));
}
}
}
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[347] 前 K 个高频元素
hash-table | heap
2022/04/08
class Solution {
public int[] topKFrequent(int[] nums, int k) {
Map<Integer, Integer> occurrences = new HashMap<Integer, Integer>();
for (int num : nums) {
occurrences.put(num, occurrences.getOrDefault(num, 0) + 1);
}
PriorityQueue<int[]> queue = new PriorityQueue<int[]>(new Comparator<int[]>() {
public int compare(int[] m, int[] n) {
return m[1] - n[1];
}
});
for (Map.Entry<Integer, Integer> entry : occurrences.entrySet()) {
int num = entry.getKey(), count = entry.getValue();
if (queue.size() == k) {
if (queue.peek()[1] < count) {
queue.poll();
queue.offer(new int[] { num, count });
}
} else {
queue.offer(new int[] { num, count });
}
}
int[] ret = new int[k];
for (int i = 0; i < k; ++i) {
ret[i] = queue.poll()[0];
}
return ret;
}
}
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[394] 字符串解码
stack | depth-first-search
2022/04/08
class Solution {
public String decodeString(String s) {
StringBuffer ans = new StringBuffer(s);
Stack<int[]> leftParenthesis = new Stack<>();
for (int i = 0; i < ans.length(); i++) {
if (ans.charAt(i) == '[') {
int num = 0, k = 1, j = i - 1;
for (; j >= 0; j--) {
if (ans.charAt(j) < '0' || ans.charAt(j) > '9') {
break;
} else {
num += k * (ans.charAt(j) - '0');
k *= 10;
}
}
int[] tmp = { j + 1, i, num };
leftParenthesis.add(tmp);
} else if (ans.charAt(i) == ']') {
int[] last = leftParenthesis.pop();
StringBuffer decodedString = new StringBuffer();
for (int j = 0; j < last[2]; j++) {
decodedString.append(ans.substring(last[1] + 1, i));
}
ans.replace(last[0], i + 1, decodedString.toString());
i = last[0] + decodedString.length() - 1;
}
}
return ans.toString();
}
}
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Hard
[301] 删除无效的括号
depth-first-search | breadth-first-search
2022/04/06
class Solution {
public List<String> removeInvalidParentheses(String s) {
List<String> ans = new ArrayList<>();
Set<String> bfs = new HashSet<>();
bfs.add(s);
while (true) {
for (String str : bfs) {
if (isValid(str)) {
ans.add(str);
}
}
if (!ans.isEmpty()) {
return ans;
}
Set<String> nextBfs = new HashSet<>();
for (String str : bfs) {
for (int i = 0; i < str.length(); i++) {
if (i > 0 && str.charAt(i) == str.charAt(i - 1)) {
continue;
} else if (str.charAt(i) == '(' || str.charAt(i) == ')') {
nextBfs.add(str.substring(0, i) + str.substring(i + 1));
}
}
}
bfs = nextBfs;
}
}
private Boolean isValid(String str) {
int count = 0;
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == '(') {
count++;
} else if (str.charAt(i) == ')') {
count--;
if (count < 0) {
return false;
}
}
}
return count == 0;
}
}
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[312] 戳气球
divide-and-conquer | dynamic-programming
2022/04/05
class Solution {
public int maxCoins(int[] nums) {
int n = nums.length;
int[][] dp = new int[n + 2][n + 2];
int[] val = new int[n + 2];
val[0] = val[n + 1] = 1;
for (int i = 1; i <= n; i++) {
val[i] = nums[i - 1];
}
for (int i = n - 1; i >= 0; i--) {
for (int j = i + 2; j <= n + 1; j++) {
for (int k = i + 1; k < j; k++) {
int sum = val[i] * val[k] * val[j];
sum += dp[i][k] + dp[k][j];
dp[i][j] = Math.max(dp[i][j], sum);
}
}
}
return dp[0][n + 1];
}
}
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