上岸算法LeetCode Weekly Contest 257解题报告

【 NO.1 统计特殊四元组】

解题思路
签到题,枚举即可。

代码展示

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[c] == nums[d]) {
                       res++;
                  }
              }
          }
      }
  }
   return res;

}
}

【 NO.2 游戏中弱角色的数量】

解题思路
按照攻击力、防御力从小到大排序,然后逆序统计即可。要注意处理攻击力相同的情况。

代码展示

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 表示大于 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 访问完所有房间的第一天】

解题思路
动态规划,dp[i] 表示访问完第 i 个房间的最小天数。

代码展示

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
只要元素之间有公因数,那么他们就可以任意排序。所以我们将有相同公因数的元素排序,最后再看序列整体是否有序即可。

代码展示

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> sets() {

   Map> 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> 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;

}
}

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