LeetCode 第282题

LeetCode 第282题
很机智

一开始以为结果中每个数只会有一位，提交了之后才发现，每个数的位数也不限制。
最大的亮点，dfs函数的int lastNum和int lastSymbol参数，略机智。
lastNum表示上一位的数，lastSymbol表示上一位的符号
假如目前的表达式为1+2*3，接下来要与4进行加减乘，那么这次递归的lastNum为6，lastSymbol为+，这样才能使计算结果正确。同时要区分lastSymbol为减号和lastSymbol为加号、乘号的区别

class Solution {
    List resList = new ArrayList<>();
    public List addOperators(String num, int target) {
        if(String.valueOf(target).equals(num)){
            resList.add(num);
        }
        StringBuilder sb = new StringBuilder();
        for (int i = 1; i < num.length(); i++){
            String sub = num.substring(0, i);
            if(sub.startsWith("0") && sub.length() > 1){
                break;
            }
            sb.append(sub);
            int temp = Integer.parseInt(sub);
            dfs(num, target, i, temp, sb, temp, 0);
            sb.delete(sb.length()-sub.length(), sb.length());
        }
        return resList;
    }

    public void dfs(String num, int target, int index, long sum, StringBuilder sb, int lastNum, int lastSymbol){
        if(index >= num.length()){
            if(sum == target){
                resList.add(sb.toString());
            }
            return;
        }

        for(int i = index+1; i < num.length()+1; i++){
            String sub = num.substring(index, i);
            if(sub.startsWith("0") && sub.length() > 1){
                break;
            }
            int temp = Integer.parseInt(sub);
            sb.append("+").append(sub);
            dfs(num, target, i, sum+temp, sb, temp, 0);
            sb.delete(sb.length()-sub.length()-1, sb.length());

            sb.append("-").append(sub);
            dfs(num, target, i, sum-temp, sb, temp, 1);
            sb.delete(sb.length()-sub.length()-1, sb.length());

            sb.append("*").append(sub);
            if(lastSymbol == 1){
                dfs(num, target, i, (sum-(long)lastNum*(long)(temp-1)), sb, temp*lastNum, lastSymbol);
            }else{
                dfs(num, target, i, (sum+(long)lastNum*(long)(temp-1)), sb, temp*lastNum, lastSymbol);
            }
            sb.delete(sb.length()-sub.length()-1, sb.length());
        }
    }
}

核心代码

if(lastSymbol == 1){
                dfs(num, target, i, (sum-(long)lastNum*(long)(temp-1)), sb, temp*lastNum, lastSymbol);
            }else{
                dfs(num, target, i, (sum+(long)lastNum*(long)(temp-1)), sb, temp*lastNum, lastSymbol);
            }

其他的都是简单的递归，这两行是解决问题，保证计算结果正确的核心