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  • LeetCode-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 -> []
    

    Credits:
    Special thanks to @davidtan1890 for adding this problem and creating all test cases.

    Analysis:

    Use recursion+backtracking. We need deal with

    1. Numbers cannot start with 0

    2. When do multiply, say previous value is x = a+b*c, in current layer, we get number d and do multiplying, the value should be x-b*c+b*c*d. So we need to carry the information of the last mulitplied number into next layer, i.e., b*c*d.

     1 public class Solution {
     2     public List<String> addOperators(String num, int target) {
     3         List<String> resList = new ArrayList<String>();
     4         StringBuilder builder = new StringBuilder();
     5         addOperRecur(resList, num, target, builder, 0, 0, 0);
     6         return resList;
     7     }
     8 
     9     public void addOperRecur(List<String> resList, String num, int target, StringBuilder builder, int pos, long value,
    10             long lastMultipledNum) {
    11         if (pos >= num.length() && value == target) {
    12             resList.add(builder.toString());
    13             return;
    14         }
    15 
    16         int builderLen = builder.length();
    17         for (int i = pos; i < num.length(); i++) {
    18             // only allow "0", not "0XXX".
    19             if (i > pos && num.charAt(pos) == '0')
    20                 return;
    21             long curNum = Long.parseLong(num.substring(pos, i + 1));
    22             if (pos == 0) {
    23                 builder.append(curNum);
    24                 addOperRecur(resList, num, target, builder, i + 1, curNum, curNum);
    25                 builder.setLength(builderLen);
    26             } else {
    27                 builder.append('+').append(curNum);
    28                 addOperRecur(resList, num, target, builder, i + 1, value + curNum, curNum);
    29                 builder.setLength(builderLen);
    30 
    31                 builder.append('-').append(curNum);
    32                 addOperRecur(resList, num, target, builder, i + 1, value - curNum, -curNum);
    33                 builder.setLength(builderLen);
    34 
    35                 builder.append('*').append(curNum);
    36                 addOperRecur(resList, num, target, builder, i + 1, value - lastMultipledNum + lastMultipledNum * curNum,
    37                         lastMultipledNum * curNum);
    38                 builder.setLength(builderLen);
    39             }
    40         }
    41     }
    42 }
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  • 原文地址:https://www.cnblogs.com/lishiblog/p/5824410.html
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