【LBLD】二维数组的花式遍历技巧

【LBLD】二维数组的花式遍历技巧

151. 反转字符串中的单词

• 思路：
  • 反转整个字符串
  • 然后反转每个单词

class Solution {
public:
    string reversePartString(string s, int a, int b) {
        if (a < 0 || b >= s.size()) {
            cout << "索引错误！" << endl;
        }
        int p1 = a, p2 = b;
        char c = 0;
        while (p1 < p2) {
            c = s[p1];
            s[p1] = s[p2];
            s[p2] = c;
            p1++; p2--;
        }
        return s;
    }

    string reverseWords(string s) {
        // 清除空格
        int p1 = 0, p2 = 0;
        while (p2 < s.size()) {
            if (s[p2] != ' '
                || (p2-1 >= 0 && p2+1 < s.size()) && s[p2-1] != ' ') {
                s[p1] = s[p2];
                p1++;
                p2++;
            }
            else while (p2 < s.size() && s[p2] == ' ') {
                p2 ++;
            }
        }
        if (s[p1-1] == ' ') p1--;
        while (p1 < s.size()) {
            s.erase(p1);
        }
        // 反转整个字符串
        s = reversePartString(s, 0, s.size()-1);
        // 反转每个单词
        p1 = 0; p2 = 0;
        while (p2 < s.size()) {
            while (s[p2] != ' ' && p2 < s.size()) {
                p2++;
            }
            s = reversePartString(s, p1, p2-1);
            p1 = ++p2;
        }
        return s;
    }
};

48. 旋转图像

class Solution {
public:
    void rotate(vector<vector<int>>& matrix) {
        // 对角
        int temp = -1;
        for (int a=0; a<matrix.size(); a++) {
            for (int b=a; b<matrix.size(); b++) {
                temp = matrix[a][b];
                matrix[a][b] = matrix[b][a];
                matrix[b][a] = temp;
            }
        }
        // 垂直
        int p1 = 0, p2 = matrix.size()-1;
        for (int a=0; a<matrix.size(); a++) {
            p1 = 0; p2 = matrix.size()-1;
            while (p1 < p2) {
                temp = matrix[a][p1];
                matrix[a][p1] = matrix[a][p2];
                matrix[a][p2] = temp;
                p1++; p2--;
            }
        }
    }
};

54. 螺旋矩阵

class Solution {
public:
    vector<int> spiralOrder(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        int left_bound = -1, right_bound = n;
        int upper_bound = -1, lower_bound = m;
        vector<int> res;
        while (res.size() < m * n) {
            // 从左向右
            for (int i=left_bound+1; i<right_bound; i++) {
                res.push_back(matrix[upper_bound+1][i]);
            }
            upper_bound++;
            if (upper_bound+1 >= lower_bound) break;
            // 从上向下
            for (int i=upper_bound+1; i<lower_bound; i++) {
                res.push_back(matrix[i][right_bound-1]);
            }
            right_bound--;
            if (left_bound+1 >= right_bound) break;
            // 从右向左
            for (int i=right_bound-1; i>left_bound; i--) {
                res.push_back(matrix[lower_bound-1][i]);
            }
            lower_bound--;
            if (upper_bound+1 >= lower_bound) break;
            // 从下向上
            for (int i=lower_bound-1; i>upper_bound; i--) {
                res.push_back(matrix[i][left_bound+1]);
            }
            left_bound++;
            if (left_bound+1 >= right_bound) break;
        }
        return res;
    }
};

59. 螺旋矩阵 II

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector<vector<int>> res(n, vector<int>(n, 0));
        int i = 1;
        int i_bound = n*n;
        int left = 0, right = n-1;
        int upper = 0, lower = n-1;
        while (i <= i_bound) {
            for (int j=left; j<=right; j++) {
                res[upper][j] = i;
                i++;
            }
            upper++;
            if (i > i_bound) break;
            for (int j=upper; j<=lower; j++) {
                res[j][right] = i;
                i++;
            }
            right--;
            for (int j=right; j>=left; j--) {
                res[lower][j] = i;
                i++;
            }
            lower--;
            for (int j=lower; j>=upper; j--) {
                res[j][left] = i;
                i++;
            }
            left++;
        }
        return res;
    }
};