2022-11-05：给定一个逆波兰式，转化成正确的中序表达式。要求只有必要加括号的地方才加括号。

答案2022-11-05：

代码用rust编写。代码如下：

计算结果。用栈。遇到数字，入栈；遇到运算符，出栈。

给出表达式，难点在于去掉没必要的小括号。准备两个栈，一个栈存数字，另一个栈存类型，类型有数字类型，加减类型，乘除类型。最有可能加括号的一定是后来的乘除类型遇到先来的加减类型。

执行结果如下：

fn main() {    let rpn = "3 -5 13 + * 6 2 3 - 2 + / + 4 5 3 * * -";    println!("{}", get_ans(rpn));    println!("{}", convert(rpn));}// 请保证给定的逆波兰式是正确的！fn get_ans(rpn: &str) -> i32 {    if rpn == "" {        return 0;    }    let parts: Vec<&str> = rpn.split(" ").collect();    let mut stack: Vec<i32> = vec![];    for part in parts.iter() {        if *part == "+" || *part == "-" || *part == "*" || *part == "/" {            let right = stack.pop().unwrap();            let left = stack.pop().unwrap();            let mut ans: i32 = 0;            if *part == "+" {                ans = left + right;            } else if *part == "-" {                ans = left - right;            } else if *part == "*" {                ans = left * right;            } else {                ans = left / right;            }            stack.push(ans);        } else {            let a: i32 = part.parse().unwrap();            stack.push(a);        }    }    // stack 只有一个数，最终的结果    return stack.pop().unwrap();}#[derive(PartialEq)]enum Operation {    SingleNumber,    AddOrMinus,    MultiplyOrDivide,}// 请保证输入的逆波兰式是正确的// 否则该函数不保证正确性// 逆波兰式仅支持+、-、*、/// 想支持更多算术运算符自己改fn convert(rpn: &str) -> String {    if rpn == "" {        return String::from("");    }    let parts: Vec<&str> = rpn.split(" ").collect();    let mut stack1: Vec<String> = vec![];    let mut stack2: Vec<Operation> = vec![];    for cur in parts.iter() {        // cur 当前遇到的字符串        // +- */ 单数        if *cur == "+" || *cur == "-" {            let b = stack1.pop().unwrap();            let a = stack1.pop().unwrap();            stack2.pop();            stack2.pop();            stack1.push(format!("{}{}{}", a, cur, b));            stack2.push(Operation::AddOrMinus);        } else if *cur == "*" || *cur == "/" {            let b = stack1.pop().unwrap();            let a = stack1.pop().unwrap();            let b_op = stack2.pop().unwrap();            let a_op = stack2.pop().unwrap();            let left = if a_op == Operation::AddOrMinus {                format!("({})", a)            } else {                a            };            let right = if b_op == Operation::AddOrMinus {                format!("({})", b)            } else {                b            };            stack1.push(format!("{}{}{}", left, cur, right));            stack2.push(Operation::MultiplyOrDivide);        } else {            stack1.push(String::from(*cur));            stack2.push(Operation::SingleNumber);        }    }    return String::from(stack1.pop().unwrap());}

执行结果如下：

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[左神java代码]()