Expression Add Operators

最新推荐文章于 2020-11-05 00:11:14 发布

sjphiChina

最新推荐文章于 2020-11-05 00:11:14 发布

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分类专栏：刷遍数

本文链接：https://blog.youkuaiyun.com/sjphiChina/article/details/52267355

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本文介绍了一种递归方法来解决字符串中添加运算符的问题，目的是找出所有可能的运算组合并计算其结果等于目标值的情况。该算法适用于解决数学表达式的生成与求解问题。

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public class Solution {
    public List<String> addOperators(String num, int target) {
        List<String> res = new LinkedList<>();
        helper(res, num, target, "", 0, 0);
        return res;
    }
    
    private void helper(List<String> res, String num, int target, String expression, long curRes, long prevRes) {
        if (num.length() == 0) {
            if (curRes == target) {
                res.add(expression);
            }
            return;
        }
        for (int i = 1; i <= num.length(); i++) {
            String curStr = num.substring(0, i);
            if (curStr.length() > 1 && curStr.charAt(0) == '0') {
                return;
            }
            long curNum = Long.valueOf(curStr);
            String nextNum = num.substring(i);
            if (expression.length() > 0) {
                helper(res, nextNum, target, expression + "*" + curStr, curRes - prevRes + prevRes*curNum, prevRes*curNum);
                helper(res, nextNum, target, expression + "+" + curStr, curRes + curNum, curNum);
                helper(res, nextNum, target, expression + "-" + curStr, curRes - curNum, -curNum);
            } else {
                helper(res, nextNum, target, curStr, curNum, curNum);
            }
        }
    }
}