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对于一个类，我必须编写一个处理表格的程序。我做了困难的部分，但现在我需要为它写一个算法复杂度分析。我基本上了解这是什么，我可以将其应用于循环等简单的事情，但教授想要对整个代码进行分析。我该怎么做呢？

第一的：

// CARTESIAN PRODUCT
    // headers first
    for (int i = 0; i < relation1[0].length; i++) {
        cartProd[0][i] = relation1[0][i];
    }
    for (int i = 0; i < relation2[0].length; i++) {
        cartProd[0][(i + relation1[0].length)] = relation2[0][i];
    }
    // Merge rows, but skip the first row of everything (headers)
    for (int z = 1; z < relation1.length; z++) {
        for (int i = 1; i < relation2.length; i++) {
            for (int j = 0; j < relation2[0].length + relation1[0].length; j++) {
                int cartRow = (i + ((relation2.length - 1) * (z - 1)));
                if (j < relation1[0].length) {
                    cartProd[cartRow][j] = relation1[z][j];
                } else {
                    cartProd[cartRow][j] = relation2[i][j - relation1[0].length];
                }
            }
        }
    }
    System.out.println("\nCartesian Product: \n"
            + Arrays.deepToString(cartProd).replaceAll("],", "]," + System.getProperty("line.separator")));

第二：

        for (int i = 1, colCount = 0; i < relation1.length; i++) {
        for (int j = 0; j < relation1[0].length; j++) {
            if (relation1[0][j].equals(commonColumnName)) {
                commonWord = relation1[i][j];
                for (int x = 0; x < relation1[0].length; x++) {
                    if (CurrentColumnValid(relation1[0][x], header_natural)) {
                        naturalJoinString += ((relation1[i][x]) + " "); 
                        if(i==1) {
                            myNaturalHeaders[colCount]=relation1[0][x]; colCount++;
                        }
                    }
                }
                // Right
                for (int a = 0; a < relation2.length; a++) {
                    for (int b = 0; b < relation2[0].length; b++) {
                        if (relation2[0][b].equals(commonColumnName) && relation2[a][b].equals(commonWord)) {
                            for (int c = 0; c < relation2[0].length; c++) {
                                if (CurrentColumnValid(relation2[0][c], header_natural)
                                        && (!relation2[a][c].equals(commonWord))) {
                                    naturalJoinString += ((relation2[a][c]) + " ");
                                    if(i==1) {
                                        myNaturalHeaders[colCount]=relation2[0][c]; colCount++;
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }

问题是，我很确定我把它弄得尽可能复杂。据我所知，我编写程序的方式是防弹测试的，所以我想保留它。但是，我不知道如何为整个程序编写复杂性分析，更不用说四个 for 循环了：/

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