227. Basic Calculator II
题目:
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.
链接: http://leetcode.com/problems/basic-calculator-ii/
题解:
比较简练的方法就是把String根据运算符先split一下,之后再计算各部分,另外也可以用Token来做。自己做的方法比较繁琐,就是从头到尾遍历String,然后根据运算符的不同做一些特殊的处理。这里几个小地方,比如一开始 s += '+'来处理最后的元素,保存之前的运算符,之前运算的结果等等。二刷的时候一定要好好优化。
Time Complexity - O(n), Space Complexity - O(1)
public class Solution { public int calculate(String s) { if(s == null || s.length() == 0) return 0; s += '+'; char prevOperator = '+'; int res = 0, prevNum = 0, curNum = 0; for(int i = 0; i < s.length(); i++) { char c = s.charAt(i); if(c == ' ') continue; else if (Character.isDigit(c)) curNum = curNum * 10 + (int)(c - '0'); else if (c == '+' || c == '-') { if(prevOperator == '+' || prevOperator == '-') { if(prevOperator == '+') { res += prevNum + curNum; } else { res += prevNum - curNum; } } else { if(prevOperator == '*') res += prevNum * curNum; else res += prevNum / curNum; } curNum = 0; prevNum = 0; prevOperator = c; } else if (c == '*' || c == '/') { if(prevOperator == '+') prevNum = curNum; else if(prevOperator == '-') prevNum = -curNum; else if(prevOperator == '*') prevNum *= curNum; else if(prevOperator == '/') prevNum /= curNum; curNum = 0; prevOperator = c; } } return res; } }
Reference:
https://leetcode.com/discuss/41790/10-16-lines-java-easy
https://leetcode.com/discuss/41558/20ms-o-n-time-o-1-space-one-scan-c-solution
https://leetcode.com/discuss/41641/17-lines-c-easy-20-ms
https://leetcode.com/discuss/41902/share-my-java-solution
https://leetcode.com/discuss/42903/java-straight-forward-iteration-solution-with-comments-stack
https://leetcode.com/discuss/42423/28ms-code-with-stacks-for-oprand-extension-cover-also-given