LeetCode 第 14 场双周赛

基础的 api 还是不够熟悉啊

5112. 十六进制魔术数字

class Solution {
 public:
  char *lltoa(long long num, char *str, int radix) {
    const char index[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
    unsigned long long unum;
    int i = 0, j, k;
    if(radix == 10 && num < 0) {
      unum = (unsigned) - num;
      str[i++] = '-';
    } else
      unum = (unsigned long long)num;
    do {
      str[i++] = index[unum % (unsigned)radix];
      unum /= radix;
    } while(unum);
    str[i] = '\0';

    if(str[0] == '-')
      k = 1;
    else
      k = 0;
    char temp;
    for(j = k; j <= (i - k - 1) / 2.0; j++) {
      temp = str[j];
      str[j] = str[i - j - 1];
      str[i - j - 1] = temp;
    }
    return str;
  }

  string toHexspeak(string num) {
    long long val = atoll(num.c_str());
    char str[1024];
    lltoa(val, str, 16);
    num = string(str);
    //cout <<val <<"\n"<< str << "\n" <<num << "\n";
    for(int i = 0, sz = num.length(); i < sz; ++i) {
      if(num[i] == '0')
        num[i] = 'O';
      else if(num[i] == '1')
        num[i] = 'I';
    }
    string ans = num;
    for(int i = 0, sz = num.length(); i < sz; ++i) {
      if(num[i] >= 'A' && num[i] <= 'Z')
        continue;
      ans = "ERROR";
    }
    return ans;
  }
};

一时没有想起 atoll 、 sprintf 等函数，写得很暴力。 long long 16 进制输出可用 %llX 。

class Solution {
 public:
  string toHexspeak(string num) {
    long long val = atoll(num.c_str());
    char str[100];
    sprintf(str, "%llX", val);
    for(int i = 0, sz = strlen(str); i < sz; ++i)
      if(str[i] == '1')
        str[i] = 'I';
      else if(str[i] == '0')
        str[i] = 'O';
      else if(str[i] > '0' && str[i] <= '9') {
        string ans("ERROR");
        return ans;
      }
    string ans = string(str);
    return ans;
  }
};

5113. 删除区间

留意区间交集是一个点的情况。

#define PB push_back
class Solution {
 public:
  vector<vector<int>> removeInterval(vector<vector<int>>& intervals, vector<int>& toBeRemoved) {
    vector<vector<int>> ans;
    for(vector<int> interval : intervals) {
      if(interval[0] >= toBeRemoved[1] || (interval[1] <= toBeRemoved[0])) {
        ans.PB(interval);
      } else if(toBeRemoved[0] < interval[0]) {
        if(toBeRemoved[1] >= interval[1])
          continue;
        else if(toBeRemoved[1] < interval[1])
          ans.PB(vector<int> {toBeRemoved[1], interval[1]});
      } else if(toBeRemoved[0] < interval[1]) {
        if(toBeRemoved[1] < interval[1]) {
          if(interval[0] != toBeRemoved[0])
            ans.PB(vector<int> {interval[0], toBeRemoved[0]});
          if(toBeRemoved[1] != interval[1])
            ans.PB(vector<int> {toBeRemoved[1], interval[1]});
        } else if(toBeRemoved[1] >= interval[1]) {
          if(interval[0] != toBeRemoved[0])
            ans.PB(vector<int> {interval[0], toBeRemoved[0]});
        }
      }
    }
    return ans;
  }
};

5114. 删除树节点

#define PB push_back

typedef pair<int, int> PII;

class Solution {
 public:
  static const int maxn = 1e4 + 7;
  vector<int> g[maxn];
  int deleteTreeNodes(int nodes, vector<int>& parent, vector<int>& value) {
    int root(-1);
    for(int i = 0; i < nodes; ++i) {
      if(parent[i] == -1)
        root = i;
      else
        g[parent[i]].PB(i);
    }
    return DFS(value, root).second;
  }

  PII DFS(vector<int>&value, int root) {
    PII ans({value[root], 1}); //{values,size}
    for(int u : g[root]) {
      PII sub = DFS(value, u);
      if(sub.first != 0)
        ans.first += sub.first, ans.second += sub.second;
    }
    return ans;
  }

};

5136. 矩形内船只的数目

划分成 4 块或者 块+线 的情况。

/**
 * // This is Sea's API interface.
 * // You should not implement it, or speculate about its implementation
 * class Sea {
 *   public:
 *     bool hasShips(vector<int> topRight, vector<int> bottomLeft);
 * };
 */

class Solution {
 public:

  int countShips(Sea sea, vector<int> topRight, vector<int> bottomLeft) {
    int ans(0);
    if(topRight[0] == bottomLeft[0] && topRight[1] == bottomLeft[1]) {
      return sea.hasShips(topRight, bottomLeft) ? 1 : 0;
    } else if(!sea.hasShips(topRight, bottomLeft)) {
      return 0;
    }
    if(bottomLeft[0] < topRight[0] && bottomLeft[1] < topRight[1]) {
      int mx = (topRight[0] + bottomLeft[0]) / 2, my = (topRight[1] + bottomLeft[1]) / 2;
      ans = countShips(sea, vector<int> {mx, my}, bottomLeft)
            + countShips(sea, vector<int> {mx, topRight[1]}, vector<int> {bottomLeft[0], my + 1})
            + countShips(sea, topRight, vector<int> {mx + 1, my + 1})
            + countShips(sea, vector<int> {topRight[0], my}, vector<int> {mx + 1, bottomLeft[1]});

    } else if(bottomLeft[0] == topRight[0]) {
      int my = (topRight[1] + bottomLeft[1]) / 2;
      ans += countShips(sea, vector<int> {bottomLeft[0], my}, bottomLeft)
      + countShips(sea, topRight, vector<int> {bottomLeft[0], my + 1});
    } else if(bottomLeft[1] == topRight[1]) {
      int mx = (topRight[0] + bottomLeft[0]) / 2;
      ans = countShips(sea, vector<int> {mx, bottomLeft[1]}, bottomLeft)
      + countShips(sea, topRight, vector<int> {mx + 1, bottomLeft[1]});
    }
    return ans;
  }
};

原文地址：https://www.cnblogs.com/Forgenvueory/p/11964864.html