Fall and can not keep only a minimum of two original array, hacked.

More than 25% of the number of elements in the array 1287. orderly appearance

To give you a non-decreasing order integer array, this array is known exactly an integer number of times it occurs more than 25% of the total number of array elements.

Please find and return the integer

Example:

**输入：** arr = [1,2,2,6,6,6,6,7,10]
**输出：** 6

Tip Hint

prompt:

1 <= arr.length <= 10^4
0 <= arr[i] <= 10^5

Code

class Solution {
public:
    int findSpecialInteger(vector<int>& arr) {
        int n = arr.size(),val = arr[0];
        pair<int,int>ans({0,0});
        for(int i = 0,c = 0;i < n;++i){
          if(arr[i] == val)  c++,ans=max(ans,make_pair(c,val));
          else c = 1,val = arr[i];
        }
        return ans.second;
    }
};

1288. deleted covered range

Give you a range list, you delete the interval list are covered by other sections.

Only if c <= aand b <= dwhen we believe that the interval [a,b)is the interval [c,d)covered.

After completing all the delete operation, you return to the list of the number of remaining range.

Example:

**输入：** intervals = [[1,4],[3,6],[2,8]]
**输出：** 2
**解释：** 区间 [3,6] 被区间 [2,8] 覆盖，所以它被删除了。

Tip Hint

Tips:

1 <= intervals.length <= 1000
0 <= intervals[i][0] < intervals[i][1] <= 10^5
For all i != j:intervals[i] != intervals[j]

Code

class Solution {
  typedef pair<int, int>PII;
 public:

  static bool cmp(vector<int>& a, vector<int>& b) {
    return (a[0] != b[0]) ? (a[0] < b[0]) : (a[1] > b[1]);
  }

  int removeCoveredIntervals(vector<vector<int>>& intervals) {
    const int n = intervals.size();
    sort(intervals.begin(), intervals.end(), cmp);
    int l = intervals[0][1], ans = 1;
    for(int i = 1; i < n; ++i) {
      if(intervals[i][1] > l)
        ans++, l = intervals[i][1];
    }
    return ans;
  }
};

1286. letter combinations iterator

You design an iterator class, including the following:

A constructor function, the input parameters comprising: a unique and ordered character string characters(the string contains only lowercase letters) and a number combinationLength.
Function Next () , according to the dictionary order to return the length of combinationLengththe next letter combination.
Function the hasNext () , there is only the length of combinationLengththe next letter combination, before returning True; otherwise, return False.

Example:

CombinationIterator iterator = new CombinationIterator("abc", 2); // 创建迭代器 iterator

iterator.next(); // 返回 "ab"
iterator.hasNext(); // 返回 true
iterator.next(); // 返回 "ac"
iterator.hasNext(); // 返回 true
iterator.next(); // 返回 "bc"
iterator.hasNext(); // 返回 false

Tip Hint

prompt:

1 <= combinationLength <= characters.length <= 15
Each test contains up to 10^4twice the function call.
Topic ensure that each function calls nextare when the presence of a combination of letters.

Code

class CombinationIterator {
 public:
  string characters;
  string cur;
  int combinationLength;
  bool first;
  CombinationIterator(string characters, int combinationLength) {
    this->characters = characters;
    this->combinationLength = combinationLength;
    this->cur = characters.substr(0, combinationLength);
    this->first = true;
  }

  string next() {
    if(first) {
      first = false;
      return cur;
    }
    vector<int>pos;
    for(int i = 0; i < combinationLength; i++) {
      pos.push_back(characters.find_first_of(cur[i]));
    }
    for(int i = pos.size() - 1; i >= 0; --i) {
      if((i == pos.size() - 1 && pos[i] < characters.length() - 1)
          || (i < pos.size() - 1 && pos[i] + 1 != pos[i + 1])) {
        shiftPos(pos, i);
        break;
      }
    }
    cur = "";
    for(int i = 0, sz = pos.size(); i < sz; ++i)
      cur += characters[pos[i]];
    return cur;
  }

  void shiftPos(vector<int>&pos, int i) {
    pos[i]++;
    for(int j = i + 1; j < combinationLength; ++j)
      pos[j] = pos[j - 1] + 1;
  }

  bool hasNext() {
    vector<int>pos;
    for(int i = 0; i < combinationLength; i++) {
      pos.push_back(characters.find_first_of(cur[i]));
    }
    //for(auto i : pos) cout << i << " "; cout << endl;
    if(pos[0] == characters.length() - combinationLength)
      return false;
    return true;
  }
};

/**
 * Your CombinationIterator object will be instantiated and called as such:
 * CombinationIterator* obj = new CombinationIterator(characters, combinationLength);
 * string param_1 = obj->next();
 * bool param_2 = obj->hasNext();
 */

1289. Minimum descending path and II

Give you a square integer arr, the definition of "non-zero offset decline path" as: from arrselecting a numeric array in each row, and the order of elected figures, the neighbors are not in the same column of the original array.

Please return to a non-zero offset of the minimum number and descending path.

Sample input and output sample Sample Input and Sample Output

Example 1:

**输入：** arr = [[1,2,3],[4,5,6],[7,8,9]]
**输出：** 13
**解释：**
所有非零偏移下降路径包括：
[1,5,9], [1,5,7], [1,6,7], [1,6,8],
[2,4,8], [2,4,9], [2,6,7], [2,6,8],
[3,4,8], [3,4,9], [3,5,7], [3,5,9]
下降路径中数字和最小的是 [1,5,7] ，所以答案是 13 。

Tip Hint

prompt:

1 <= arr.length == arr[i].length <= 200
-99 <= arr[i][j] <= 99

Code

class Solution {
 public:
  static const int inf = 0x3f3f3f3f;
  int minFallingPathSum(vector<vector<int>>& arr) {
    int n = arr.size(), m = arr[0].size();
    int s[n][m];
    for(int j = 0; j < m; ++j)
      s[0][j] = arr[0][j];
    for(int i = 1; i < n; ++i) {
      for(int j = 0; j < m; ++j) {
        s[i][j] = inf;
        for(int k = 0; k < m; ++k) {
          if(k == j) continue;
          s[i][j] = min(s[i][j], s[i - 1][k] + arr[i][j]);
        }
      }
    }
    int ans = inf;
    for(int i = 0;i < m;++i) ans = min(ans,s[n-1][i]);
    return ans;
  }
};

LeetCode Game 15 Game Fortnight

More than 25% of the number of elements in the array 1287. orderly appearance

Tip Hint

Code

1288. deleted covered range

Tip Hint

Code

1286. letter combinations iterator

Tip Hint

Code

1289. Minimum descending path and II

Sample input and output sample Sample Input and Sample Output

Tip Hint

Code

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