共计 2503 个字符，预计需要花费 7 分钟才能阅读完成。
【NO.1 统计非凡四元组】
解题思路
签到题，枚举即可。
代码展现
class Solution {
public int countQuadruplets(int[] nums) {
int n = nums.length;
int res = 0;
for (int a = 0; a < n; a++) {for (int b = a + 1; b < n; b++) {for (int c = b + 1; c < n; c++) {for (int d = c + 1; d < n; d++) {if (nums[a] + nums[b] + nums == nums[d]) {res++;}
}
}
}
}
return res;}
}
【NO.2 游戏中弱角色的数量】
解题思路
依照攻击力、防御力从小到大排序，而后逆序统计即可。要留神解决攻击力雷同的状况。
代码展现
class Solution {
public int numberOfWeakCharacters(int[][] properties) {
Arrays.sort(properties, (a, b) -> {if (a[0] == b[0]) {return a[1] - b[1];
}
return a[0] - b[0];
});
int res = 0;
int lastAttack = properties[properties.length - 1][0];
int lastDefense = properties[properties.length - 1][1];
int maxDefense = 0; // maxDefense 示意大于 lastAttack 的角色中，最大的防御力
for (int i = properties.length - 2; i >= 0; i--) {if (properties[i][0] < lastAttack) {maxDefense = Math.max(maxDefense, lastDefense);
lastAttack = properties[i][0];
lastDefense = properties[i][1];
}
if (properties[i][1] < maxDefense) {res++;}
}
return res;}
}
【NO.3 拜访完所有房间的第一天】
解题思路
动静布局，dp[i] 示意拜访完第 i 个房间的最小天数。
代码展现
class Solution {
public int firstDayBeenInAllRooms(int[] nextVisit) {
int n = nextVisit.length;
long[] dp = new long[n];
long P = (long) (1e9 + 7);
for (int i = 1; i < n; ++i) {dp[i] = (2 * dp[i - 1] - dp[nextVisit[i - 1]] + 2 + P) % P;
}
return (int) dp[n - 1];}
}
【NO.4 数组的最大公因数排序】
解题思路
只有元素之间有公因数，那么他们就能够任意排序。所以咱们将有雷同公因数的元素排序，最初再看序列整体是否有序即可。
代码展现
class UnionFind {
public UnionFind(int size) {
f = new int[size];
Arrays.fill(f, -1);}
public int find(int x) {
if (f[x] < 0)
return x;
return f[x] = find(f[x]);}
public boolean merge(int a, int b) {
int fa = find(a);
int fb = find(b);
if (fa == fb)
return false;
f[fa] = fb;
return true;}
public Map<Integer, List<Integer>> sets() {
Map<Integer, List<Integer>> res = new HashMap<>();
for (int i = 0; i < f.length; i++) {int fi = find(i);
if (!res.containsKey(fi)) {res.put(fi, new ArrayList<>());
}
res.get(fi).add(i);
}
return res;}
private int[] f;
}
class Solution {
public boolean gcdSort(int[] nums) {
Map<Integer, List<Integer>> set = new HashMap<>();
for (int i = 0; i < nums.length; i++) {for (int j = 1; j * j <= nums[i]; j++) {if (nums[i] % j == 0) {if (!set.containsKey(j)) {set.put(j, new ArrayList<>());
}
set.get(j).add(i);
if (j * j < nums[i]) {int k = nums[i] / j;
if (!set.containsKey(k)) {set.put(k, new ArrayList<>());
}
set.get(k).add(i);
}
}
}
}
UnionFind uf = new UnionFind(nums.length);
for (var e : set.entrySet()) {if (e.getKey() < 2) {continue;}
var list = e.getValue();
for (int i = 1; i < list.size(); i++) {uf.merge(list.get(i - 1), list.get(i));
}
}
var sets = uf.sets();
int[] res = new int[nums.length];
for (var e : sets.entrySet()) {var list = e.getValue();
var sortedList = new ArrayList<>(list);
sortedList.sort(Comparator.comparingInt(a -> nums[a]));
for (int i = 0; i < list.size(); i++) {res[list.get(i)] = nums[sortedList.get(i)];
}
}
for (int i = 1; i < res.length; i++) {if (res[i] < res[i - 1]) {return false;}
}
return true;}
}
正文完
