剑指Offer常见问题整理

1 在一个二维数组中，每一行都按照从左到右递增的顺序排序，每一列都按照从上到下递增的顺序排序。请完成一个函数，输入这样的一个二维数组和一个整数，判断数组中是否含有该整数。（来自牛客网，剑指offer）

#include <iostream>
#include <vector>
std::vector<std::vector<int>> s;
void init(int n) {
    int pos = n*2;
    for (int i = 0; i < n; ++i) {
        std::vector<int> tmp;
        for (int j = pos; j < pos + n; ++j) {
            tmp.push_back(j);
        }
        pos = pos + n;
        s.push_back(tmp);
        tmp.clear();
    }
}

void print(std::vector<std::vector<int>> &s, int n) {
    for (int i=0; i < n; ++i) {
        for (auto itr:s[i]) {
            std::cout << itr << ",";
        }
        std::cout << "\n";
    }
}

std::pair<int, int> find(std::vector<std::vector<int>> &s, int dst) {
    int line_cnt = s[0].size();
    int col_cnt = s.size();

    int low = 0;
    int high = col_cnt - 1;
    int nMid = (low + high) / 2;
    // 可能的列
    int sus_col_pos = -1;
    while (low <= high) {
        nMid = (low + high) / 2;
        if (s[nMid][0] <= dst && s[nMid][s[nMid].size()-1] >= dst) {
            sus_col_pos = nMid;
            break;
        } else if (s[nMid][s[nMid].size()-1] < dst) {
            low = nMid + 1;
        } else {
            high = nMid - 1;
        }
    }

    //std::cout<< sus_col_pos;
    // 二分找到了列， 对行进行二分查找
    low = 0;
    high = s[sus_col_pos].size()-1;
    int dst_pos = -1;
    while (low <= high) {
        int nMid = (low + high) / 2;
        if (s[sus_col_pos][nMid] == dst) {
            dst_pos = nMid;
            break;
        } else {
            if (s[sus_col_pos][nMid] > dst) {
                high = nMid - 1;
            } else {
                low = nMid + 1;
            }
        }
    }
    return std::make_pair(sus_col_pos, dst_pos);
}

int bi_search(int a[], int low, int high, int dst) {
    int nMid = (low + high) / 2;
    while(low <= high) {
        nMid = (low + high) / 2;
        if (a[nMid] ==  dst) {
            return nMid;
        } else {
            if (a[nMid] > dst) {
                high = nMid - 1;
            } else {
                low = nMid + 1;
            }
        }
    }
    return nMid;
}

std::ostream &operator<< (std::ostream &out, std::pair<int, int> &s) {
    //std::cout << "(" << s.first << "," << s.second << ")" << "\n";
    out << "(" << s.first << "," << s.second << ")" << "\n";
    return out;
}
int main() {
    init(5);
    print(s, 5);
    std::pair<int, int> res = find(s, 34);
    std::cout << res;
    // std::cout << find(s, 20).first;
//    int a[] = {2, 4, 6, 34, 90};
//    std::cout << bi_search(a, 0, 4, 5);
    return 0;
}

　　

请实现一个函数，将一个字符串中的空格替换成“ % 20”。例如，当字符串为We Are Happy.则经过替换之后的字符串为We%20Are%20Happy。
//注意如果输出的是%20d需要对%进行转义

//用Stl中的vector时间复杂度为：O(str.length());空间复杂度O(str.length+3*SpaceNum)

void ReplaceSpace( string strSrc,char *sOut)
{ 
    vector<char> cOut;
    const char *pStr = strSrc.data();
    while (*pStr != '\0')
    {
        if (*pStr == ' ')
        {
            cOut.push_back('%');
            cOut.push_back('2');
            cOut.push_back('0');
        }
        else
            cOut.push_back(*pStr);
        pStr++;
    }
    cOut.push_back('\0');

    for (int i = 0; i < cOut.size(); i++)
    {
        sOut[i] = cOut[i];
    }

}

//Test
　　string str= "ni hao ma";
    char pStr[32] = {0};
    ReplaceSpace(str,pStr);
    printf("%s",pStr);
    getchar();
    return 0;