[随缘一题]后缀表达式问题

来源:

维基百科-后缀表达式

目标

将中缀表达式转换为后缀表达式,比如((5+2) * (8-3))/4 转换为5 2 + 8 3 - * 4 /.

解题思路

将表达式的字符逐一处理,如果是数字(变量)则直接输出,如果是字符入栈,并按以下规则进行处理.

+/-: 低优先级,所以将栈中的所有运算符出栈,之后将自己入栈.

*or/:高优先级,将栈中的其他乘除运算符出栈,之后将自己入栈.

(: 左括号则直接入栈.

): 右括号将栈中运算符逐一出栈,直到遇到左括号.

实现代码

/**
 * solve the N-Queen problem
 */
public class NQueen {

  //the number of chess board,example 8
  private static final int N = 8;

  // result, the result[i] mean: the location of [i] line is on result[i] column.
  private int[] result = new int[N];

  //total num of possible result
  private int resultNum = 0;

  /**
   * calculation
   */
  private void calculation(int n) {

    //if n == N, print the result
    if (n == N) {
      for (int i = 0; i < result.length; i++) {
        System.out.print(result[i] + ",");
      }
      System.out.println();
      resultNum++;
    } else {
      for (int i = 0; i < N; i++) {
        // test every location possible
        result[n] = i;
        //if line n is allowed, locate the next line
        if (isAllowed(n)) {
          calculation(n + 1);
        }
      }
    }
  }

  /**
   * judge current line is allowed or not.
   */
  private boolean isAllowed(int i) {
    // i is not allowed while it in same line or diagonal with the pre line
    for (int j = 0; j < i; j++) {
      if (result[i] == result[j] || Math.abs(i - j) == Math.abs(result[i] - result[j])) {
        return false;
      }
    }
    return true;
  }

  //main method, include some test cases
  public static void main(String[] args) {
    NQueen queen = new NQueen();

    queen.calculation(0);

    System.out.println(queen.resultNum);
  }

}

完。

ChangeLog

2019-02-24 完成

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