227. Basic Calculator II

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

一刷
题解:
用3个变量,preVal, curVal, sign
如果碰到了数字,就保存为curVal, 并且与当前已有的sign,每次保存sign再get下一个number
如果sign是乘除,那么直接计算preVal sign curVal, 变为新的preVal
如果sign是加减,将当前res+preVal,preVal = sign*curVal

public class Solution {
    public int calculate(String s) {
        if (s == null) return 0;
        s = s.trim().replaceAll(" +", "");//remove all the space
        int length = s.length();
    
        int res = 0;
        long preVal = 0; // initial preVal is 0
        char sign = '+'; // initial sign is +
        int i = 0;
        while(i<length){
            long curVal = 0;
            while (i < length && Character.isDigit(s.charAt(i))) { // int
                curVal = curVal*10 + (s.charAt(i) - '0');
                i++;
            }
            if (sign == '+') {
                res += preVal;  // update res
                preVal = curVal;
            } else if (sign == '-') {
                res += preVal;  // update res
                preVal = -curVal;
            } else if (sign == '*') {
                preVal = preVal * curVal; // not update res, combine preVal & curVal and keep loop
            } else if (sign == '/') {
                preVal = preVal / curVal; // not update res, combine preVal & curVal and keep loop
            }
            if (i < length) { // getting new sign
                sign = s.charAt(i);
                i++;
            }
        }
        res += preVal;
        return res;
    }
}

二刷
题解:这个题目的特点是没有括号。用res保存乘除运算之前的值即可。不需要stack.
preVal sign curVal,
如果sign为加减,则res+=preVal, preVal = sign*curVal
如果sign为乘除,res+= preVal sign curVal

public class Solution {
    public int calculate(String s) {
        long preVal = 0, curVal = 0;
        s = s.trim().replaceAll(" +", "");
        int len = s.length();
        int res = 0;
        char sign = '+';
        int i=0;
        while(i<len){
            curVal = 0;
            while(i<len && Character.isDigit(s.charAt(i))){
                curVal = curVal*10 + s.charAt(i) - '0';
                i++;
            }
            if(sign == '+'){
                res += preVal;
                preVal = curVal;
            }else if(sign == '-'){
                res += preVal;
                preVal = -curVal;
            }else if(sign == '*'){
                res += preVal * curVal;
            }else if(sign == '/'){
                res += preVal / curVal;
            }
            if(i<len){
                sign = s.charAt(i);
                i++;
            }
        }
        return res;
    }
}
最后编辑于
©著作权归作者所有,转载或内容合作请联系作者
【社区内容提示】社区部分内容疑似由AI辅助生成,浏览时请结合常识与多方信息审慎甄别。
平台声明:文章内容(如有图片或视频亦包括在内)由作者上传并发布,文章内容仅代表作者本人观点,简书系信息发布平台,仅提供信息存储服务。

推荐阅读更多精彩内容

友情链接更多精彩内容