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题目：
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 -> []题解：
注意的问题：
1. 有可能string很长，所以要用长整数。
2.需要记录last oprand，为的是处理乘号的情况。乘号时的计算公式为: result - lastOP + lastOP * curVaule，如 2 + 3 * 5, 当计算到要乘5时，result = 2 + 3 = 5, lastOp = 3, curValue = 5, 则最终结果为： 5 - 3 + 3 * 5 = 17 = 2 + 3 * 5。
3. 记录当前结果，用于在num长度为零时， 判断是否与target相等，并加入最终的results中。
4. 需要跳过长于一个零的操作数，如“00”， “000”， etc.
5. 按左操作数从长度为1 到长度为len(num)循环， 并递归。
C++版：
class Solution {
public:vector<string> addOperators(string num, int target) {vector<string> results;opRecur(num, target, 0, 0, "", results);return results;}void opRecur(string num, int target, long long lastOp, long long result, string expression, vector<string> & results) {if(num.length() == 0 && result == target) {results.push_back(expression);return;}for(int i = 1; i <= num.length(); i++) {string cur = num.substr(0, i);string rest = num.substr(i);long long curV = stoll(cur);if(cur.length() > 1 && cur[0] == '0')continue;if(expression.length() == 0) {opRecur(rest, target, curV, curV, cur, results);} else {opRecur(rest, target, curV, result + curV, expression + "+" + cur, results);opRecur(rest, target, -curV, result - curV, expression + "-" + cur, results);opRecur(rest, target, lastOp * curV, result - lastOp + lastOp * curV, expression + "*" + cur, results);}}}
};Java版：
public class Solution {public List<String> addOperators(String num, int target) {List<String> results = new ArrayList<>();opRecur(num, target, 0, 0, "", results);return results;}private void opRecur(String num, int target, long lastOp, long result, String expression, List<String> results) {if(num.length() == 0) {if(target == result)results.add(expression);return;} for(int i = 1; i <= num.length(); i++) {String cur = num.substring(0, i);if(cur.length() > 1 && cur.charAt(0) == '0')continue;String rest = num.substring(i);long curV = Long.parseLong(cur);if(expression.length() == 0) {opRecur(rest, target, curV, curV, expression + cur, results);} else {opRecur(rest, target, curV, result + curV, expression + "+" + cur, results);opRecur(rest, target, -curV, result - curV, expression + "-" + cur, results);opRecur(rest, target, curV * lastOp, result - lastOp + lastOp * curV, expression + "*" + cur, results);}}}
}Python版：
class Solution(object):def addOperators(self, num, target):""":type num: str:type target: int:rtype: List[str]"""results = []self.opRecur(num, target, 0, 0, "", results)return resultsdef opRecur(self, num, target, lastOp, result, expression, results):if len(num) == 0:if target == result:results.append(expression)returnfor i in range(1, len(num) + 1):cur = num[:i]if len(cur) > 1 and cur[0] == "0":continuerest = num[i:]curV = int(cur)if len(expression) == 0:self.opRecur(rest, target, curV, curV, expression + cur, results)else:self.opRecur(rest, target, curV, result + curV, expression + "+" + cur, results)self.opRecur(rest, target, -curV, result - curV, expression + "-" + cur, results)self.opRecur(rest, target, lastOp * curV, result - lastOp + lastOp * curV, expression + "*" + cur, results)这篇关于LeetCode 题解(211) : Expression Add Operators的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
