上岸算法LeetCode Weekly Contest 259解题报告

【 NO.1 执行操作后的变量值】

解题思路
签到题。

class Solution {
public int finalValueAfterOperations(String[] operations) {

   int v = 0;
   for (String op : operations) {
       if (op.contains("++")) {
           v++;
      } else {
           v--;
      }
  }
   return v;

【 NO.2 数组美丽值求和】

解题思路
由前缀最大值和后缀最小值即可得到中间元素的美丽值，所以预处理出前缀最大值和后缀最小值数组即可。

class Solution {
public int sumOfBeauties(int[] nums) {

   int[] preMax = new int[nums.length];
   preMax[0] = nums[0];
   for (int i = 1; i < nums.length; i++) {
       preMax[i] = Math.max(preMax[i - 1], nums[i]);
  }
   int[] sufMin = new int[nums.length];
   sufMin[nums.length - 1] = nums[nums.length - 1];
   for (int i = nums.length - 2; i >= 0; i--) {
       sufMin[i] = Math.min(sufMin[i + 1], nums[i]);
  }
   int res = 0;
   for (int i = 1; i < nums.length - 1; ++i) {
       if (preMax[i - 1] < nums[i] && nums[i] < sufMin[i + 1]) {
           res += 2;
      } else if (nums[i - 1] < nums[i] && nums[i] < nums[i + 1]) {
           res += 1;
      }
  }
   return res;

【 NO.3 检测正方形】

解题思路
使用 Map 储存所有的顶点，然后在 count 查询时枚举对角线。

class DetectSquares {

Map<Integer, Integer> count;

public DetectSquares() {

   count = new HashMap<>();

public void add(int[] point) {

   int c = comp(point[0], point[1]);
   count.put(c, count.getOrDefault(c, 0) + 1);

public int count(int[] point) {

   int res = 0;
   for (var kv : count.entrySet()) {
       int x = X(kv.getKey());
       int y = Y(kv.getKey());
       if (Math.abs(x - point[0]) == Math.abs(y - point[1]) && x != point[0]) {
           res += kv.getValue() *
                   count.getOrDefault(comp(x, point[1]), 0) *
                   count.getOrDefault(comp(point[0], y), 0);
      }
  }
   return res;

private int comp(int x, int y) {

   return x * 10000 + y;

private int X(int c) {

   return c / 10000;

private int Y(int c) {

   return c % 10000;

【 NO.4 重复 K 次的最长子序列】

解题思路
注意 2 <= n < k * 8，而如果一个子序列想要重复出现 k 次，那么这个子序列中的每个字符都至少要出现 k 次，所以说答案的长度一定小于等于 7。

我们首先找出来所有出现次数不小于 k 次的字符，然后枚举这些字符的排列组合，依次判断每一个排列组合是否出现了 k 次。

class Solution {
public String longestSubsequenceRepeatedK(String s, int k) {

   Map<Character, Integer> count = new HashMap<>();
   for (char c : s.toCharArray()) {
       count.put(c, count.getOrDefault(c, 0) + 1);
  }
   StringBuilder s2 = new StringBuilder();
   for (char c : s.toCharArray()) {
       if (count.get(c) >= k) {
           s2.append(c);
      }
  }
   count.clear();
   for (char c : s2.toString().toCharArray()) {
       count.put(c, count.getOrDefault(c, 0) + 1);
  }
   return solve(new StringBuilder(), count, s2.toString().toCharArray(), k);

private String solve(StringBuilder cur, Map<Character, Integer> count, char[] s, int k) {

   String res = "";
   var keys = new HashSet<Character>(count.keySet());
   for (var c : keys) {
       cur.append(c);
       if (comp(cur.toString(), res)) {
           int cnt = 0, idx = 0;
           for (char cc : s) {
               if (cc == cur.charAt(idx) && ++idx == cur.length()) {
                   idx = 0;
                   if (++cnt == k) {
                       res = cur.toString();
                       break;
                  }
              }
          }
      }
       int bak = count.get(c);
       if (bak - k < k) {
           count.remove(c);
      } else {
           count.put(c, bak - k);
      }
       String r = solve(cur, count, s, k);
       if (comp(r, res)) {
           res = r;
      }
       cur.deleteCharAt(cur.length() - 1);
       count.put(c, bak);
  }
   return res;

private boolean comp(String a, String b) {

   return a.length() > b.length() || (a.length() == b.length() && a.compareTo(b) > 0);