The fourth question of Java Blue Bridge Cup 2014

Java Reverse Polish Expression

A normal expression is called an infix expression. The operator is in the middle, which is mainly for human reading, and it is not convenient for machines to solve it.

Example: 3 + 5 * (2 + 6) - 1

Also, parentheses are often required to change the order of operations.

Conversely, if the reverse Polish expression (prefix expression) is used, the above expression is expressed as:

• • 3 * 5 + 2 6 1

Parentheses are no longer needed, and the machine can easily solve it recursively.

For simplicity, we assume:

1. There are only + - * three operators
2. Each operand is a non-negative integer less than 10

The following program evaluates a reverse Polish representation string.
Its return value is an array: the first element is the result of the evaluation, and the second is the number of characters it has parsed.

static int[] evaluate(String x) { if (x.length() == 0) return new int[] { 0, 0 };

    char c = x.charAt(0);
    if (c >= '0' && c <= '9')
        return new int[] { c - '0', 1 };

    int[] v1 = evaluate(x.substring(1));
    int[] v2 = __________________________________________; // 填空位置

    int v = Integer.MAX_VALUE;
    if (c == '+')
        v = v1[0] + v2[0];
    if (c == '*')
        v = v1[0] * v2[0];
    if (c == '-')
        v = v1[0] - v2[0];

    return new int[] { v, 1 + v1[1] + v2[1] };
}

public static void main(String[] args) {
    
    
		// TODO Auto-generated method stub
		int t[] = evaluate("-+3*5+261");
		System.out.println(t[0]);
	}
 
	static int[] evaluate(String x) {
    
    
		// 3 * 5 + 2 6 1
		// + 3 * 5 + 2 6 1
		// - + 3 * 5 + 2 6 1
		if (x.length() == 0)
			return new int[] {
    
     0, 0 };
 
		char c = x.charAt(0);
		if (c >= '0' && c <= '9')
			return new int[] {
    
     c - '0', 1 };
 
		int[] v1 = evaluate(x.substring(1));
		// [3][1]
		int[] v2 = evaluate(x.substring(v1[1] + 1)); // 填空位置
 
		int v = Integer.MAX_VALUE;
		if (c == '+')
			v = v1[0] + v2[0];
		if (c == '*')
			v = v1[0] * v2[0];
		if (c == '-')
			v = v1[0] - v2[0];
 
		return new int[] {
    
     v, 1 + v1[1] + v2[1] };
	}

Fill in the blanks: evaluate(x.substring(v1[1] + 1));