Expression Add Operators
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Given a string that contains only digits 0-9
and a target value, return all possibilities to addbinary 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 -> []要打起精神来!!!
这个是典型的DFS,有思路,也考虑到溢出,但是还是没有做出来。
【解题思路】
1、典型DFS,可以搜索出所有答案。
2、因为不止是一位数字,所以每次需要考虑多位数字组合在一起的情况。
3、注意如果截取的时候,首位为0,就不需要处理了,可能得到错误答案。
4、注意截取多位数据类型为long的情况。因为num为string型,int可能溢出。
public class Solution { public List<String> addOperators(String num, int target) { List<String> list = new ArrayList<String>(); dfs(list, num, "", 0, 0, 0, target); return list; } private void dfs(List<String> list, String num, String path, int pos, long sum, long lastNum, int target) { int len = num.length(); if (pos == len && sum == target) { list.add(path); return; } if (pos >= len) { return; } for (int i = pos; i < len; i++) { if (i != pos && num.charAt(pos) == '0') { break; } long curNum = Long.parseLong(num.substring(pos, i + 1)); if (pos == 0) { dfs(list, num, path + "" + curNum, i + 1, curNum, curNum, target); } else { dfs(list, num, path + "+" + curNum, i + 1, sum + curNum, curNum, target); dfs(list, num, path + "-" + curNum, i + 1, sum - curNum, -curNum, target); dfs(list, num, path + "*" + curNum, i + 1, sum - lastNum + lastNum * curNum, lastNum * curNum, target); } } } }参考链接:http://blog.csdn.net/u013027996/article/details/48713751
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