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Implement atoi to convert a string to an integer.
Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.
Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.
Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.
spoilers alert... click to show requirements for atoi.
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class Solution { public: /* * 这个问题需要考虑很多边界情况 * 1.如果str为空,返回空 * 2.如果str前面有空格,过滤掉空格 * 3.如果str > INT_MAX ,返回 INT_MAX * 4.如果str < INT_MIN, 返回 INT_MIN * 5.如果str格式不匹配,返回已经匹配成功的部分 */ int myAtoi(string str) { int begin = 0; int end = str.length() - 1; //过滤掉左右的空格 while (str[begin] == ‘ ‘) { begin++; } while (str[end] == ‘ ‘) { end--; } //设置负号 bool minus = false; if (‘-‘ == str[begin]) { minus = true; begin++; } else if (‘+‘ == str[begin]) { minus = false; begin++; } double res = 0; for (begin; begin<=end; begin++) { if (str[begin] >= ‘0‘ && str[begin] <= ‘9‘) { res = res * 10 + str[begin] - ‘0‘; } else { break; } } if (!minus) { if (res > INT_MAX) { res = INT_MAX; } } else { res = res * -1; if (res < INT_MIN) { res = INT_MIN; } } return (int)res; } };
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原文地址：http://www.cnblogs.com/SpeakSoftlyLove/p/5097125.html
