共计 3589 个字符，预计需要花费 9 分钟才能阅读完成。
【NO.1 字符串转化后的各位数字之和】
解题思路
循环 k 次加和即可。
代码展现
class Solution {
public int getLucky(String s, int k) {
String s2 = "";
for (int i = 0; i < s.length(); i++) {s2 += String.valueOf((int) (s.charAt(i) - 'a' + 1));
}
int result = 0;
for (int i = 0; i < k; i++) {
result = 0;
for (int j = 0; j < s2.length(); j++) {result += s2.charAt(j) - '0';
}
s2 = String.valueOf(result);
}
return result;
}}
【NO.2 子字符串渐变后可能失去的最大整数】
解题思路
贪婪，这个子串的终点要尽可能得靠左，并且渐变后肯定要变大。
代码展现
class Solution {
public String maximumNumber(String num, int[] change) {
StringBuilder result = new StringBuilder();
int status = 0;
for (int i = 0; i < num.length(); i++) {char oldChar = num.charAt(i);
char newChar = (char) (change[oldChar - '0'] + '0');
if (status == 2) { // status == 2 示意曾经完结了渐变，间接应用 oldChar
result.append(oldChar);
} else if (status == 1) { // status == 1 示意正在渐变，进行比照，决定是否完结渐变
if (oldChar <= newChar) {result.append(newChar);
} else {result.append(oldChar);
status = 2;
}
} else if (status == 0) { // status == 0 示意还没开始渐变，进行比照，决定是否开始渐变
if (oldChar < newChar) {result.append(newChar);
status = 1;
} else {result.append(oldChar);
}
}
}
return result.toString();}
}
【NO.3 最大兼容性评分和】
解题思路
回溯，枚举所有的可能性即可。
代码展现
class Solution {
int max;
public int maxCompatibilitySum(int[][] students, int[][] mentors) {
max = 0;
boolean[] vis = new boolean[mentors.length];
int[][] compat = new int[students.length][mentors.length];
for (int i = 0; i < students.length; i++) {for (int j = 0; j < mentors.length; j++) {for (int k = 0; k < students[0].length; k++) {if (students[i][k] == mentors[j][k]) {compat[i][j]++;
}
}
}
}
dfs(0, 0, compat, students.length, students[0].length, vis);
return max;}
void dfs(int stu, int sum, int[][] compat, int n, int m, boolean[] vis) {
max = Math.max(max, sum);
// 剪枝优化：若前面的学生每对儿都是最大匹配度，也不迭以后的最优解，则不必要再持续递归
if (stu == n || sum + (n - stu) * m <= max) {return;}
for (int i = 0; i < n; i++) {if (!vis[i]) {vis[i] = true;
dfs(stu + 1, sum + compat[stu][i], compat, n, m, vis);
vis[i] = false;
}
}}
}
【NO.4 删除零碎中的反复文件夹】
解题思路
Hash
文件目录零碎是树结构，为每棵子树计算哈希值，最初将哈希值雷同的子树删掉即可。
计算哈希的办法比拟多，最简略的能够间接转换成 JSON 字符串，然而效率略低。能够利用子节点的哈希值计算以后节点的哈希值，效率较高。
代码展现
class Solution {
static class Node {
boolean deleted;
int hash;
TreeMap<String, Node> children = new TreeMap<>();}
private final Node root;
private final Map<String, Integer> hash;
private final Map<Integer, Integer> count;
public Solution() {
root = new Node();
hash = new HashMap<>();
count = new HashMap<>();}
public void add(List<String> word) {
Node node = root;
for (String i : word) {if (!node.children.containsKey(i)) {Node child = new Node();
node.children.put(i, child);
}
node = node.children.get(i);
}}
private void calcHash(Node node) {
if (node.children.size() == 0) {
node.hash = 0;
return;
}
StringBuilder sb = new StringBuilder();
for (var child : node.children.navigableKeySet()) {Node childNode = node.children.get(child);
calcHash(childNode);
if (sb.length() != 0) {sb.append("/");
}
sb.append(childNode.hash);
sb.append(getHash(child));
}
node.hash = getHash(sb.toString());
count.put(node.hash, count.getOrDefault(node.hash, 0) + 1);}
private int getHash(String child) {
if (!hash.containsKey(child)) {hash.put(child, hash.size() + 1);
}
return hash.get(child);}
private void delete(Node node) {
for (var child : node.children.entrySet()) {delete(child.getValue());
}
if (count.getOrDefault(node.hash, 0) > 1) {node.deleted = true;}}
private List<List<String>> toList(Node node) {
List<List<String>> result = new LinkedList<>();
if (node != root) {result.add(new LinkedList<>());
}
if (node.children.size() == 0) {return result;}
for (var child : node.children.entrySet()) {if (child.getValue().deleted) {continue;}
List<List<String>> childList = toList(child.getValue());
for (var l : childList) {((LinkedList<String>) l).addFirst(child.getKey());
result.add(l);
}
}
return result;}
public List<List<String>> deleteDuplicateFolder(List<List<String>> paths) {
for (var path : paths) {add(path);
}
calcHash(root);
delete(root);
return toList(root);}
}
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