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Commit 0d528cc

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‎8. String to Integer (atoi).java

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/*
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习题:
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Implement atoi to convert a string to an integer.
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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.
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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.
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*/
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/*解析
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本题考察的东西就是看你思考的全不全面,而且他还没有说具体的规则,我说一下它判定的规则,分为几种情况
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1.空串
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2.数字之前有空格如:" 123"
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3.带符号的字符串 如:"+123"
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4.在数字中间夹杂字符或者空格 如:" —12 33" 将要输出 -12 后面舍弃
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5.存在溢出问题,这个我在写的时候就很迷茫。下面主要说明这个问题。
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因为是int类型的,所以占4个字节,记住 最大的最小的是 -(2^31)-1 ~ 2^31 -1(有一位是符号位,还有一个数字是0)
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判断溢出的条件就是,有两种可能,1.结果>最大位/10 2.结果=最大位/10 并且 digit>最大位%10 只有这两种可能,因为一个数位的变化就是增加或者减少10。
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代码如下:
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*/
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class Solution {
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public int myAtoi(String str) {
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int sybol=1;//符号 0正数 1负数
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int result = 0;
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//先判断第一个字符
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//分情况讨论:
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//只有"+""-"
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//空""
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//非数字,
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//纯数字
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if(str==null) {
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return 0;
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}
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if(str.charAt(0)=='-'||str.charAt(0)=='+')
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if(str.length()==1)
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return 0;
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else if(str.charAt(0)=='-')
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sybol=0;
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str=str.sustring(1);
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}
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return stringToInt(str,sybol);
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}
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public int stringToInt(String str,int sybol) {
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int digit;
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int result=0;
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//逐一扫描
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while(int i = 0;i<str.length();i++) {
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digit = str.charAt(i);
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//如果是数字的话
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if(digit>='0'||digit<='9') {
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result = result*10 + digit;
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}
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else {
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return 0;
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}
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}
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if(sybol==0){
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result = -result;
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}
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}
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return result;
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}

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