题目大意：把一串数字分成两段，输出所有合法的分法（可以加小数点）。

https://leetcode.com/problems/ambiguous-coordinates/description/

We had some 2-dimensional coordinates, like "(1, 3)" or "(2, 0.5)".  Then, we removed all commas, decimal points, and spaces, and ended up with the string S.  Return a list of strings representing all possibilities for what our original coordinates could have been.

Our original representation never had extraneous zeroes, so we never started with numbers like “00”, “0.0”, “0.00”, “1.0”, “001”, “00.01”, or any other number that can be represented with less digits.  Also, a decimal point within a number never occurs without at least one digit occuring before it, so we never started with numbers like “.1”.

The final answer list can be returned in any order.  Also note that all coordinates in the final answer have exactly one space between them (occurring after the comma.)

Example 1:
Input: "(123)"
Output: ["(1, 23)", "(12, 3)", "(1.2, 3)", "(1, 2.3)"]

Example 2:
Input: "(00011)"
Output:  ["(0.001, 1)", "(0, 0.011)"]
Explanation: 
0.0, 00, 0001 or 00.01 are not allowed.

Example 3:
Input: "(0123)"
Output: ["(0, 123)", "(0, 12.3)", "(0, 1.23)", "(0.1, 23)", "(0.1, 2.3)", "(0.12, 3)"]

Example 4:
Input: "(100)"
Output: [(10, 0)]
Explanation: 
1.0 is not allowed.

Solution: Brute Force

Time complexity: O(n^3)

Space complexity: O(n)

C++

// Author: Huahua
// Running time: 16 ms
class Solution {
public:
  vector<string> ambiguousCoordinates(string S) {
    vector<string> ans;
    int n = S.length();
    for (int l1 = 1; l1 <= n - 3; ++l1) {      
      int l2 = n - l1 - 2;
      string s1 = S.substr(1, l1);
      string s2 = S.substr(l1 + 1, l2);      
      if (valid(s1) && valid(s2))
        ans.push_back("(" + s1 + ", " + s2 + ")");
      
      for (int i = 0; i < l1; ++i) {
        string ts1 = s1;
        if (i > 0) ts1.insert(i, ".");        
        if (!valid(ts1)) continue;
        for (int j = 0; j < l2; ++j) {
          if (i == 0 && j == 0) continue;
          string ts2 = s2;
          if (j > 0) ts2.insert(j, ".");          
          if (!valid(ts2)) continue;          
          ans.push_back("(" + ts1 + ", " + ts2 + ")");
        }
      }
    }
    
    return ans;
  }
private:    
  bool valid(const string& s) {
    bool p = false; // has "."
    bool d = false; // has digits in front of "."
    int zeros = 0;  // leading zeros
    int i = 0;
    while (s[zeros] == '0' && zeros < s.length()) ++zeros;
    for (int i = zeros; i < s.length(); ++i) {
      if (!p && isdigit(s[i])) d = true;
      if (s[i] == '.') p = true;
    }
    if (zeros == 1 && s != "0" && !p || zeros > 1) return false;
    if (p && (s.back() == '0' || s.back() == '.' || zeros > 0 && d)) return false;    
    return true;
  }
};

The post 花花酱 LeetCode 816. Ambiguous Coordinates appeared first on Huahua's Tech Road.

Problem:

Given an array containing n distinct numbers taken from 0, 1, 2, …, n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

Idea:

sum / xor

Solution:

// Author: Huahua
class Solution {
public:
    // Solution 1: Sum
    int missingNumber(vector<int>& nums) {
        int n = nums.size();
        int x = (0+n)*(n+1)/2 - accumulate(nums.begin(), nums.end(), 0);
        return x;
    }
    
    // Solution 2: XOR
    int missingNumber(vector<int>& nums) {
        int n = nums.size();
        int x = 0;
        for(int i=1;i<=n;++i)
            x = x ^ i ^ nums[i-1];
        return x;
    }
};

The post 花花酱 LeetCode 268. Missing Number appeared first on Huahua's Tech Road.