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  • [LeetCode] Expression Add Operators

    Expression Add Operators

    Given a string that contains only digits 0-9 and a target value, return all possibilities to add binary operators (not unary) +-, or * between the digits so they evaluate to the target value.

    Examples: 

    "123", 6 -> ["1+2+3", "1*2*3"] 
    "232", 8 -> ["2*3+2", "2+3*2"]
    "105", 5 -> ["1*0+5","10-5"]
    "00", 0 -> ["0+0", "0-0", "0*0"]
    "3456237490", 9191 -> []

    DFS,注意*的优先级高,所以要记录当前最后一个操作数(*表达式)的值。注意整形溢出,用long long来存中间结果。

     1 class Solution {
     2 public:
     3     void dfs(vector<string> &res, string num, string out, int target, long long diff, long long curVal) {
     4         if (num.empty() && curVal == target) {
     5             res.push_back(out);
     6             return;
     7         }
     8         string curNum, nextNum;
     9         for (int i = 1; i <= num.size(); ++i) {
    10             curNum = num.substr(0, i);
    11             nextNum = num.substr(i);
    12             if (curNum.size() > 1 && curNum[0] == '0') return;
    13             if (out.empty()) {
    14                 dfs(res, nextNum, curNum, target, stoll(curNum), stoll(curNum));
    15             } else {
    16                 dfs(res, nextNum, out + "+" + curNum, target, stoll(curNum), curVal + stoll(curNum));
    17                 dfs(res, nextNum, out + "-" + curNum, target, -stoll(curNum), curVal - stoll(curNum));
    18                 dfs(res, nextNum, out + "*" + curNum, target, diff * stoll(curNum), curVal - diff + diff * stoll(curNum));
    19             }
    20         }
    21     }
    22     vector<string> addOperators(string num, int target) {
    23         vector<string> res;
    24         dfs(res, num, "", target, 0, 0);
    25         return res;
    26     }
    27 };
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  • 原文地址:https://www.cnblogs.com/easonliu/p/4829399.html
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