共计 2487 个字符，预计需要花费 7 分钟才能阅读完成。
【NO.1 反转单词前缀】
解题思路
签到题。
代码展现
class Solution {
public String reversePrefix(String word, char ch) {
int index = word.indexOf(ch);
return new StringBuffer(word.substring(0, index + 1)).reverse().toString() +
word.substring(index + 1);
}
}
【NO.2 可调换矩形的组数】
解题思路
将矩形依照长宽比分类，计数即可。
代码展现
class Solution {
static class Frac {
int den;
int num;
public static int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);
}
public Frac(int num, int den) {int g = gcd(num, den);
this.num = num / g;
this.den = den / g;
}
@Override
public boolean equals(Object o) {if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Frac frac = (Frac) o;
return den == frac.den && num == frac.num;
}
@Override
public int hashCode() {return Objects.hash(num, den);
}
}
public long interchangeableRectangles(int[][] rectangles) {
Map<Frac, Integer> count = new HashMap<>();
for (var rec : rectangles) {Frac f = new Frac(rec[0], rec[1]);
count.put(f, count.getOrDefault(f, 0) + 1);
}
long res = 0;
for (var k : count.entrySet()) {int v = k.getValue();
res += (long) v * (v - 1) / 2;
}
return res;
}
}
【NO.3 两个回文子序列长度的最大乘积】
解题思路
暴力枚举。应用二进制位示意一个子序列，枚举所有状况即可。
代码展现
class Solution {
public int maxProduct(String s) {
int len = s.length();
int res = 0;
int[] mem = new int[1 << len];
Arrays.fill(mem, -1);
for (int i = 0; i < (1 << len); i++) {for (int j = 0; j < (1 << len); j++) {if ((i & j) > 0) {continue;}
res = Math.max(res, length(s, i, mem) * length(s, j, mem));
}
}
return res;
}
private int length(String s, int bitset, int[] mem) {
if (mem[bitset] >= 0) {return mem[bitset];
}
mem[bitset] = 0;
for (int i = 0, j = s.length() - 1; i <= j; i++, j--) {while (i <= j && (bitset & (1 << i)) == 0) i++;
while (i <= j && (bitset & (1 << j)) == 0) j--;
if (!(i <= j && (bitset & (1 << i)) != 0 && (bitset & (1 << j)) != 0)) {break;}
if (s.charAt(i) == s.charAt(j)) {mem[bitset] += i == j ? 1 : 2;
} else {mem[bitset] = 0;
break;
}
}
return mem[bitset];
}
}
【NO.4 每棵子树内缺失的最小基因值】
解题思路
DFS 合并 Set 即可。然而有两个优化很重要：
- 如果子树中缺失的最大的是 x, 那么枚举查找以后树缺失的只须要从 x 开始即可，而不是 1
- 合并 Set 时由小 Set 合并到大 Set 中
代码展现
class Solution {
public int[] smallestMissingValueSubtree(int[] parents, int[] nums) {
Map<Integer, List<Integer>> children = new HashMap<>();
for (int i = 1; i < parents.length; i++) {if (!children.containsKey(parents[i])) {children.put(parents[i], new ArrayList<>());
}
children.get(parents[i]).add(i);
}
int[] ans = new int[parents.length];
dfs(0, children, nums, ans);
return ans;
}
private Set<Integer> dfs(int cur, Map<Integer, List<Integer>> children, int[] nums, int[] ans) {
Set<Integer> set = new HashSet<>();
set.add(nums[cur]);
if (!children.containsKey(cur)) {ans[cur] = nums[cur] == 1 ? 2 : 1;
return set;
}
var child = children.get(cur);
int start = 1;
for (var c : child) {var r = dfs(c, children, nums, ans);
if (r.size() > set.size()) {
Set<Integer> tmp = r;
r = set;
set = tmp;
}
set.addAll(r);
start = Math.max(start, ans);
}
while (set.contains(start)) {start++;}
ans[cur] = start;
return set;
}
}
正文完
