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.
Example 1:
Input:num ="123", target = 6
Output: ["1+2+3", "1*2*3"]
Example 2:
Input:num ="232", target = 8
Output: ["2*3+2", "2+3*2"]
Example 3:
Input:num ="105", target = 5
Output: ["1*0+5","10-5"]
Example 4:
Input:num ="00", target = 0
Output: ["0+0", "0-0", "0*0"]
Example 5:
Input:num ="3456237490", target = 9191
Output: []
分析
解法是利用递归回溯来遍历所有的可能，但是要注意一些边界情形。
public class Solution {
public List<String> addOperators(String num, int target) {
List<String> rst = new ArrayList<String>();
if(num == null || num.length() == 0) return rst;
helper(rst, "", num, target, 0, 0, 0);
return rst;
}
// eval记录当前计算结果，multed计算上次计算变化的部分，在选择乘法时会用到这个
public void helper(List<String> rst, String path, String num, int target, int pos, long eval, long multed){
if(pos == num.length()){
if(target == eval)
rst.add(path);
return;
}
for(int i = pos; i < num.length(); i++){
if(i != pos && num.charAt(pos) == '0') break; // 抛弃以0开始的数字
long cur = Long.parseLong(num.substring(pos, i + 1));
if(pos == 0){
helper(rst, path + cur, num, target, i + 1, cur, cur); // 起始数字特殊处理
}
else{
helper(rst, path + "+" + cur, num, target, i + 1, eval + cur , cur); // 对当前数字cur选择加上之前的部分
helper(rst, path + "-" + cur, num, target, i + 1, eval -cur, -cur); // 选择减
// 选择乘法要特殊处理，减去上次变化的部分，将这个变化的部分乘以当前的数字再加上去
helper(rst, path + "*" + cur, num, target, i + 1, eval - multed + multed * cur, multed * cur );
}
}
}
}