Leetcode double pointer water problem solution

392. Judging subsequences

Ideas ： Define two pointers to s and t And then determine s and t Whether the characters are the same , If it's the same then i and j meanwhile ++, If not, let t Skip this character to compare the next character , Finally, if i If it has been matched, it means s yes t The string of .

class Solution {
public:
    bool isSubsequence(string s, string t) {
        int i = 0,j = 0;
        while(i<s.size() && j<t.size())
        {
            if(s[i]==t[j])
            {
                i++;
                j++;
            }
            else
            {
                j++;
            }
        }
        return i==s.size();
    }
};

541. Reverse string II

Ideas ： First define a starting point pointer i And a counter count, The starting point plus the counter is less than size Just go ahead , Count , Count to 2k Just before the reversal k Characters , Then enter the final judgment count Count to the end , That is, the last interval is not satisfied 2k When , Judge this value and then k To 2k Between the two is still 0,k Between , Reverse .

class Solution {
    
public:
    string reverseStr(string s, int k) {
        int count = 0;
        int i = 0;
        while(i+count<s.size())
        {
            count++;
            if(count==2*k)
            {
                reverse(&s[i],&s[i+k]);
                i += count;
                count = 0;
            }
        }
        if(count>=k && count<2*k)
        {
            reverse(&s[i],&s[i+k]);
        }
        else if(count>0 && count<k)
        {
            reverse(&s[i],&s[i+count]);
        }
        return s;

    }
};

Interview questions 16.24. Sum of number pairs

Ideas ： Double pointer ： Define a left pointer i And the right pointer j, Sort the array , Then start as i and j The sum of positions is greater than target that j– If it is less than, then i++, Equal to put the logarithm into the answer array .

class Solution {
public:
    vector<vector<int>> pairSums(vector<int>& nums, int target) {
        vector<vector<int>> ans;
        int i = 0,j = nums.size()-1;
        sort(nums.begin(),nums.end());
        while(i<j)
        {
            int n = nums[i]+nums[j];
            if(n>target)
            {
                j--;
            }
            else if(n<target)
            {
                i++;
            }
            else
            {
                ans.push_back({nums[i],nums[j]});
                i++;j--;
            }
        }
        return ans;
    }
};

696. Counting binary substrings

Ideas ： Define a i The pointer , Constantly traverse the string s, Then one count Record continuous 1 perhaps 0 Number of occurrences , Then two adjacent 0 and 1 or 1 and 0 The smaller value of the number of occurrences is these two groups 0 and 1 The largest logarithm that can be composed .

class Solution {
public:
    int countBinarySubstrings(string s) {
        int num1 = 0,i = 0;
        int ans = 0,count = 0;
        while(i<s.size())
        {
            count = 0;
            char c = s[i];
            while(i<s.size() && c == s[i])
            {
                count++;
                i++;
            }
            ans+=min(num1,count);
            num1 = count;
        }
        return ans;
    }
};