leetcode 227: Basic Calculator II

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Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.

You may assume that the given expression is always valid.

Some examples:

"3+2*2" = 7
" 3/2 " = 1
" 3+5 / 2 " = 5

Note: Do not use the eval built-in library function.

[思路]

pass两遍, 第一遍, 先解决乘除, 第二遍, 做加减.

[CODE]

public class Solution {
    public int calculate(String s) {
        if(s==null || s.length()==0) return 0;
        
        LinkedList<Integer> list = new LinkedList<Integer>();
        
        for(int i=0; i<s.length(); i++) {
            char c = s.charAt(i);
            if(Character.isDigit(c)) {
                int cur = c-'0';
                while(i+1<s.length() && Character.isDigit(s.charAt(i+1))) {
                    cur = cur * 10 + s.charAt(i+1) - '0';
                    ++i;
                }
                if(!list.isEmpty() && (list.peek() == 2 || list.peek()==3)) {
                    int op = list.pop();
                    int opl = list.pop();
                    int res = 0;
                    if(op==2) res = opl * cur;
                    else res = opl / cur;
                    list.push(res);
                } else {
                    list.push(cur);
                }               
            } else if(c==' ') continue;
            else {
                switch (c) {
                    case '+': list.push(0);
                    break;
                    case '-': list.push(1);
                    break;
                    case '*': list.push(2);
                    break;
                    case '/': list.push(3);
                    break;
                    default: return -1;
                }
            }
        }
        
        if(list.isEmpty()) return 0;
        Collections.reverse(list);
        
        int res = list.poll();
        
        while(!list.isEmpty()) {
        	int op = list.poll();
        	int opr = list.poll();
        	if(op==0) res += opr;
        	else res -= opr;
        }
        return res;
    }
}

leetcode 227: Basic Calculator II

标签：

原文地址：http://blog.csdn.net/xudli/article/details/46644317