共计 2561 个字符，预计需要花费 7 分钟才能阅读完成。
【NO.1 句子中的无效单词数】
解题思路
签到题。
代码展现
class Solution {
public int countValidWords(String sentence) {
String[] words = sentence.split(" ");
int count = 0;
for (var word : words) {char[] chars = word.trim().toCharArray();
boolean invalid = false;
int index = -1;
for (int i = 0; i < chars.length && !invalid; i++) {char c = chars[i];
if ('0' <= c && c <= '9') {invalid = true;} else if (c == '-') {if (index == -1) {index = i;} else {invalid = true;}
} else if (isNotAlpha(c) && i != chars.length - 1) {invalid = true;}
}
if (invalid || index == 0 || index == chars.length - 1) {continue;}
if (index > 0 && (isNotAlpha(chars[index - 1]) || isNotAlpha(chars[index + 1]))) {continue;}
count++;
}
return count;}
boolean isNotAlpha(char c) {
return 'a' > c || c > 'z';}
}
【NO.2 下一个更大的数值均衡数】
解题思路
枚举即可。
代码展现
class Solution {
public int nextBeautifulNumber(int n) {
for (int i = n + 1; ; i++) {if (balance(i)) {return i;}
}}
private boolean balance(int num) {
int[] cnt = new int[10];
for (; num > 0; num /= 10) {cnt[num % 10]++;
}
for (int i = 0; i < 10; i++) {if (cnt[i] != 0 && cnt[i] != i) {return false;}
}
return true;}
}
【NO.3 统计最高分的节点数目】
解题思路
首先进行一次 DFS 求出每个节点的子树大小，而后进行一次 DFS 求出每个节点的分数。
留神计算分数须要用 Long 类型，防止乘法溢出。
代码展现
class Solution {
Long maxScore;
Map<Long, Integer> count;
public int countHighestScoreNodes(int[] parents) {
List<List<Integer>> children = new ArrayList<>();
for (int i = 0; i < parents.length; i++) {children.add(new ArrayList<>());
}
for (int i = 1; i < parents.length; i++) {children.get(parents[i]).add(i);
}
int[] sum = new int[parents.length];
calcSum(0, sum, children);
maxScore = 0L;
count = new HashMap<>();
calcMaxScore(0, sum, children);
return count.get(maxScore);}
private void calcMaxScore(int cur, int[] sum, List<List<Integer>> children) {
long score = 1;
for (var nxt : children.get(cur)) {score *= sum[nxt];
}
if (sum[0] - sum[cur] > 0) {score *= sum[0] - sum[cur];
}
count.put(score, count.getOrDefault(score, 0) + 1);
maxScore = Math.max(maxScore, score);
for (var nxt : children.get(cur)) {calcMaxScore(nxt, sum, children);
}}
private void calcSum(int cur, int[] sum, List<List<Integer>> children) {
sum[cur] = 1;
for (var nxt : children.get(cur)) {calcSum(nxt, sum, children);
sum[cur] += sum[nxt];
}}
}
【NO.4 并行课程 III】
解题思路
简直是树型 DP 模板题，比较简单。令 finish(i) 示意实现课程 i 的最短时间，则 finish(i) = max(finish(j)) + time[i]，其中 j 是 i 的前置课程。
代码展现
class Solution {
public int minimumTime(int n, int[][] relations, int[] time) {
List<List<Integer>> prev = new ArrayList<>();
for (int i = 0; i < n; i++) {prev.add(new ArrayList<>());
}
for (int[] rel : relations) {prev.get(rel[1] - 1).add(rel[0] - 1);
}
int[] mem = new int[n];
int result = 0;
for (int i = 0; i < n; i++) {result = Math.max(result, finish(i, prev, time, mem));
}
return result;}
private int finish(int cur, List<List<Integer>> prev, int[] time, int[] mem) {
if (mem[cur] > 0) {return mem[cur];
}
if (prev.get(cur).size() == 0) {return time[cur];
}
for (int p : prev.get(cur)) {mem[cur] = Math.max(mem[cur], finish(p, prev, time, mem) + time[cur]);
}
return mem[cur];}
}