字符串中数字子串的求和

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例如：

A-1BC--12 the number is -1,12 and the sum is 11

A-1BC--2C--D6E the number is -1,2,6 and the sum is 7

下面是代码：

////  main.cpp//  字符串中数字子串的求和////  Created by zjl on 16/8/13.//  Copyright © 2016年 zjl. All rights reserved.//  for example://  A-1BC--12    the number is -1,12 and the sum is 11#include <iostream>#include <string>using namespace std;bool isdigits(char c){ //判断字符是否是数字    if(c < '9' && c > '0')        return true;    else        return false;}int solve(string s){    if(s.size() == 0) return 0;    int sum = 0, count = 0, sum_part = 0;    bool isdigit = false; //判断是否有数字存在    int i = 0;    while( i < s.size()){        isdigit = false;        while(i < s.size() && s[i] == '-'){ //累计符号的数量            count++;            i++;        }        while(i < s.size() && isdigits(s[i])){ //若是数字，则统计局部和            isdigit = true;            sum_part = sum_part * 10 + (s[i] - '0');            i++;        }        if(isdigit){  //若前面是数字，则将其加到总和上去            if(count % 2 == 1)                sum_part *= -1;            sum += sum_part;            sum_part = 0;        }        count = 0;        ++i;    }    return sum;}int main(int argc, const char * argv[]) {    // insert code here...    string s = "A-1BC--2C--D6E";    int sum = solve(s);    cout<<sum<<endl;    return 0;}

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