Given an expression string array, return the final result of this expression Have you met this question in a real interview? Yes
Example
For the expression 2*6-(23+7)/(1+2),
input is [
"2", "*", "6", "-", "(",
"23", "+", "7", ")", "/",
(", "1", "+", "2", ")"
],
return 2 Note
The expression contains only integer, +, -, *, /, (, ).
这道题其实应该算Basic Calculator III. 参考了http://blog.csdn.net/nicaishibiantai/article/details/45740649
思路就是两个stack，一个存数字一个存符号。如果遇到数字直接存到数字stack；如果遇到符号，有几种情况：
1.当前符号比上一个符号优先级高，比如* 高于+，那么直接进栈
2.当前符号低于上一个，那么就要把所有已经在stack里面优先于当前符号的全算完，再推进当前符号
3.当前符号是“（”，直接push
4.当前符号是“）”，就要把所有“（”以前的符号全部算完
public class Solution {
/**
* @param expression: an array of strings;
* @return: an integer
*/
public int evaluateExpression(String[] expression) {
// write your code here
Stack<Integer> integers = new Stack<Integer>();
Stack<String> ops = new Stack<String>();
int i = 0;
while (i < expression.length) {
String cur = expression[i];
if (isOp(cur)) { // current string is an op
if (cur.equals("(")) ops.push(cur);
else if (cur.equals(")")) {
while (ops.size()>0 && !ops.peek().equals("(")) {
integers.push(calc(integers.pop(), integers.pop(), ops.pop()));
}
ops.pop();
}
else { // +,-,*,/
while (ops.size()>0 && precede(cur, ops.peek())) {
integers.push(calc(integers.pop(), integers.pop(), ops.pop()));
}
ops.push(cur);
}
}
else integers.push(Integer.parseInt(cur)); // current String is an integer, push to integer stack
i++;
}
while (!ops.isEmpty()) {
integers.push(calc(integers.pop(), integers.pop(), ops.pop()));
}
return integers.isEmpty()? 0 : integers.pop();
}
public boolean isOp(String input) {
if (input.equals("+") || input.equals("-") || input.equals("*")
|| input.equals("/") || input.equals("(") || input.equals(")"))
return true;
return false;
}
public int calc(int a, int b, String op) {
if (op.equals("+")) return a+b;
else if (op.equals("-")) return b-a;
else if (op.equals("*")) return a*b;
else return b/a;
}
public boolean precede(String a, String b) {
if (b.equals("*") || b.equals("/")) return true;
if (b.equals("+") || b.equals("-")) {
if (a.equals("*") || a.equals("/")) return false;
else return true;
}
return false; //case like (a+b) 到第一个+号时，+和（比应该return false
}
};