LeetCode: 227. Basic Calculator II

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227. Basic Calculator II

LeetCode - Basic Calculator II

这里在处理将字符串转成数字时，从左向右按位乘以十再相加的方法，这样就可以只遍历一遍字符串。

另外，这里还有个地方需要注意：在确定当前操作符的情况下，再计算上一次的值。所以，就需要把上一次的值保存下来。

思考题：如果数字是全体整数，又该如何实现？如果再扩容成实数呢？

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.

Example 1:

Input: "3+2*2"
Output: 7

Example 2:

Input: " 3/2 "
Output: 1

Example 3:

Input: " 3+5 / 2 "
Output: 5

Note:

You may assume that the given expression is always valid.
Do not use the eval built-in library function.

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/**
 * Runtime: 5 ms, faster than 98.92% of Java online submissions for Basic Calculator II.
 *
 * Memory Usage: 42.8 MB, less than 5.97% of Java online submissions for Basic Calculator II.
 *
 * Copy from: https://leetcode.com/problems/basic-calculator-ii/discuss/62996/Java-straight-forward-iteration-Solution-with-comments-No-Stack-O(N)-and-O(1)[Java straight forward iteration Solution with comments, No Stack, O(N) & O(1) - LeetCode Discuss]
 */
public int calculate(String s) {
    if (Objects.isNull(s) || s.length() == 0) {
        return 0;
    }
    int result = 0;
    long previous = 0;
    char operator = '+';

    int length = s.length();

    for (int i = 0; i < length; i++) {
        char c = s.charAt(i);
        if (c == ' ') {
            continue;
        }
        if (Character.isDigit(c)) {
            int value = c - '0';
            while (i + 1 < length && Character.isDigit(s.charAt(i + 1))) {
                value = value * 10 + (s.charAt(i + 1) - '0');
                i++;
            }
            if (operator == '+') {
                result += previous;
                previous = value;
            } else if (operator == '-') {
                result += previous;
                previous = -value;
            } else if (operator == '*') {
                previous *= value;
            } else if (operator == '/') {
                previous /= value;
            }
        } else {
            operator = c;
        }
    }
    result += previous;
    return result;
}