一道模拟题

给你一个字符串表达式 s ,请你实现一个基本计算器来计算并返回它的值。

注意:不允许使用任何将字符串作为数学表达式计算的内置函数,比如 eval()

示例 1:

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输入:s = "1 + 1"
输出:2

示例 2:

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输入:s = " 2-1 + 2 "
输出:3

示例 3:

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输入:s = "(1+(4+5+2)-3)+(6+8)"
输出:23

此解法包含乘除,取模,幂

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class Solution {
Map<Character, Integer> map = new HashMap<>(){{
put('-', 1);
put('+', 1);
put('*', 2);
put('/', 2);
put('%', 2);
put('^', 3);
}};
public int calculate(String s) {
s = s.replaceAll(" ", "");
char[] cs = s.toCharArray();
int n = s.length();
Deque<Integer> nums = new ArrayDeque<>();
nums.addLast(0);
Deque<Character> ops = new ArrayDeque<>();
for (int i = 0; i < n; i++) {
char c = cs[i];
if (c == '(') {
ops.addLast(c);
} else if (c == ')') {
while (!ops.isEmpty()) {
if (ops.peekLast() != '(') {
calc(nums, ops);
} else {
ops.pollLast();
break;
}
}
} else {
if (isNumber(c)) {
int u = 0;
int j = i;
while (j < n && isNumber(cs[j])) u = u * 10 + (cs[j++] - '0');
nums.addLast(u);
i = j - 1;
} else {
if (i > 0 && (cs[i - 1] == '(' || cs[i - 1] == '+' || cs[i - 1] == '-')) {
nums.addLast(0);
}
while (!ops.isEmpty() && ops.peekLast() != '(') {
char prev = ops.peekLast();
if (map.get(prev) >= map.get(c)) {
calc(nums, ops);
} else {
break;
}
}
ops.addLast(c);
}
}
}
while (!ops.isEmpty() && ops.peekLast() != '(') calc(nums, ops);
return nums.peekLast();
}
void calc(Deque<Integer> nums, Deque<Character> ops) {
if (nums.isEmpty() || nums.size() < 2) return;
if (ops.isEmpty()) return;
int b = nums.pollLast(), a = nums.pollLast();
char op = ops.pollLast();
int ans = 0;
if (op == '+') {
ans = a + b;
} else if (op == '-') {
ans = a - b;
} else if (op == '*') {
ans = a * b;
} else if (op == '/') {
ans = a / b;
} else if (op == '^') {
ans = (int)Math.pow(a, b);
} else if (op == '%') {
ans = a % b;
}
nums.addLast(ans);
}
boolean isNumber(char c) {
return Character.isDigit(c);
}
}


一道模拟题
http://example.com/2023/09/24/一道模拟题/
作者
ykexc
发布于
2023年9月24日
许可协议