C++ 解法, 执行用时: 5ms, 内存消耗: 660KB, 提交时间: 2022-07-24

class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param first int整型vector<vector<>>
     * @param second int整型vector<vector<>>
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > intersectionofintervals(vector<vector<int> >& first,
            vector<vector<int> >& second) {
        vector<vector<int> > ans;
        int p1 = 0, p2 = 0;
        while (p1 < first.size() && p2 < second.size()) {
            if (first[p1][1] < second[p2][0]) {
                ++p1;
                continue;
            }
            if (second[p2][1] < first[p1][0]) {
                ++p2;
                continue;
            }
            ans.push_back({max(first[p1][0], second[p2][0]), min(first[p1][1], second[p2][1])});
            if (first[p1][1] < second[p2][1]) {
                ++p1;
            } else {
                ++p2;
            }
        }
        return ans;
    }
};

C++ 解法, 执行用时: 5ms, 内存消耗: 660KB, 提交时间: 2022-05-19

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param first int整型vector<vector<>> 
     * @param second int整型vector<vector<>> 
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > intersectionofintervals(vector<vector<int> >& first, vector<vector<int> >& second) {
        // write code here
        vector<vector<int> > ans;
        int p1 = 0, p2 = 0;
        while (p1 < first.size() && p2 < second.size()) {
            if (first[p1][1] < second[p2][0]) {
                ++p1;
                continue;
            }
            if (second[p2][1] < first[p1][0]) {
                ++p2;
                continue;
            }
            ans.push_back({max(first[p1][0], second[p2][0]), min(first[p1][1], second[p2][1])});
            if (first[p1][1] < second[p2][1]) {
                ++p1;
            } else {
                ++p2;
            }
        }
        return ans;
    }
};

C++ 解法, 执行用时: 5ms, 内存消耗: 680KB, 提交时间: 2022-06-13

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param first int整型vector<vector<>> 
     * @param second int整型vector<vector<>> 
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > intersectionofintervals(vector<vector<int> >& first, vector<vector<int> >& second) {
        // write code here
        vector<vector<int>> result;
        int left=0,right=0;
        int n=first.size();
        int m=second.size();
        while(left<n&&right<m){
            auto f1=first[left];
            auto f2=second[right];
            if(f1[0]<=f2[0]){
                if(f2[1]<=f1[1]){
                    result.push_back({f2[0],f2[1]});
                    right++;
                }else{
                    if(f1[1]>=f2[0]){
                        result.push_back({f2[0],f1[1]});
                        left++;
                    }else{
                        left++;
                    }
                }
            }
            else{
                if(f1[1]<=f2[1]){
                    result.push_back({f1[0],f1[1]});
                    left++;
                }else{
                    if(f2[1]>=f1[0]){
                        result.push_back({f1[0],f2[1]});
                        right++;
                    }else{
                        right++;
                    }
                }
            }
        }
        return result;
    }
};

C++ 解法, 执行用时: 5ms, 内存消耗: 684KB, 提交时间: 2022-04-11

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param first int整型vector<vector<>> 
     * @param second int整型vector<vector<>> 
     * @return int整型vector<vector<>>
     */
    static bool cmp(vector<int>& a, vector<int>& b) {
        if (a[0] == b[0]) return a[1] < b[1];
        return a[0] < b[0];
    }
    
    vector<vector<int> > intersectionofintervals(vector<vector<int> >& first, vector<vector<int> >& second) {
        // write code here
        vector<vector<int>> res;
        // 1. 排序, 但是本题已经排过序，所以不用再排序
        int n1 = first.size(), n2 = second.size();
        int i = 0, j = 0;
        while (i < n1 && j < n2) {
            int a1 = first[i][0], a2 = first[i][1];
            int b1 = second[j][0], b2 = second[j][1];
            
            // a2 < b1 or b2 < a1 时没有交集 
            if (a2 >= b1 && b2 >= a1) {
                res.push_back({max(a1, b1), min(a2, b2)});
            }
            
            if (a2 < b2) {
                i++;
            } else {
                j++;
            }
        }
        return res;
    }
};

C++ 解法, 执行用时: 6ms, 内存消耗: 652KB, 提交时间: 2022-08-06

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param first int整型vector<vector<>> 
     * @param second int整型vector<vector<>> 
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > intersectionofintervals(vector<vector<int> >& first, vector<vector<int> >& second) {
        // write code here
        vector<vector<int>> ans;
        int p1=0,p2=0;
        while(p1<first.size()&&p2<second.size())
        {
            if(first[p1][1]<second[p2][0])
            {
                ++p1;
                continue;
            }
            if(second[p2][1]<first[p1][0])
            {
                ++p2;
                continue;
            }
            ans.push_back({max(first[p1][0],second[p2][0]),min(first[p1][1],second[p2][1])});
            if(first[p1][1]<second[p2][1])
            {
                ++p1;
            }else{
                ++p2;
            }
        }
        return ans;
    }
};