算法刷题记录

思路：

class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        
        int left = 0;
        int length = INT_MAX;
        int sum = 0;
        for(int right = 0; right < nums.size(); ++right){
                sum += nums[right];
                while(sum >= target && left <= right){
                    length = min(length, right - left + 1);
                    sum -= nums[left];
                    left++;
                }
        }

        return (length == INT_MAX) ? 0 : length;//正则表达式写法

    }
};

//使用vector建立哈希表

class Solution {
public:
    int maxProduct(vector<string>& words) {

        vector<vector<bool>> hash(words.size(), vector<bool>(26, false));
        for(int i = 0; i < words.size(); i++){
            for(char s1: words[i]){
                hash[i][s1 - 'a'] = true;
            }
        }

        int length = 0;
        for(int i = 0; i < words.size(); i++){
            for(int j = i + 1; j < words.size(); j++){
                int k = 0;
                for(; k < 26; k++){
                    if(hash[i][k] == true && hash[j][k] == true){
                        break;
                    }

                }
                if(k == 26){
                    int rent = words[i].size() * words[j].size();
                    length = max(length, rent);
                }
            }
        }
        return length;
    }
};

//暴力搜索法

class Solution {
public:
    int maxProduct(vector<string>& words) {
        int ret = 0;
        for (int i = 0; i < words.size(); ++i) {
            for (int j = i + 1; j < words.size(); ++j) {
                bool same = true;
                for (char ch1 : words[i]) {
                    for (char ch2 : words[j]) {
                        if (ch1 == ch2) {   
                            same = false;
                            break;
                        }
                    }
                    if (!same) {
                        break;
                    }
                }
                if (same) {
                    int cur = words[i].size() * words[j].size();
                    ret = max(ret, cur);
                }
            }
        }
        return ret;
    }
};