[LeetCode] String to Integer (atoi)

来源：互联网发布：加强网络基础设施建设编辑：程序博客网时间：2024/06/04 19:18

题目：Implement atoi to convert a string to an integer.

这道题目算法不难，不容易通过的原因是不了解 atoi 对所有类型用例的不同处理。

解思想：优先截取出最靠前的数字字符串，并判断是否有效。向上溢出输出能表示的最大值，向下溢出，输出能表示的最小值。

public class Solution {    public int myAtoi(String str) {        if(str.length() == 0)return 0;       double result = 0;char[] cha = str.toCharArray();int length = 0, start = -1, end =str.length(), flag = -1, i = 0;while(i < str.length() && (cha[i] < '0' || cha[i] > '9') ){if(cha[i] != ' ' && cha[i] != '+' && cha[i] != '-'){flag = i;break;}i++;}for( i = 0; i < str.length(); i++){if(cha[i] >= '0' && cha[i] <= '9' && start == -1)start = i;if(i >= 1 && (cha[i] < '0' || cha[i] > '9') && (cha[i - 1] >= '0' || cha[i - 1] <= '9') && cha[i - 1]!=' ' && end == str.length())end = i;}if(flag >= 0 && flag < start)return 0;if(start > 1 && (cha[start - 1] < '0' || cha[start - 1] > '9') && (cha[start - 2] < '0' || cha[start - 2] > '9') && cha[start - 2] != ' ')return 0;if(start < 0)return 0;length = end - start;if(length == 0)return 0;for( i = 0 ; i < length; i++){double tmp = (double)java.lang.Math.pow(10, (length - i - 1));result = result + (cha[i + start] - 48) * tmp;}if( start != 0 && cha[start - 1] == '-')result = 0 - result; if(result > Integer.MAX_VALUE)result = Integer.MAX_VALUE;if((result < Integer.MIN_VALUE))    result = Integer.MIN_VALUE;return (int)result;            }}

1047/1047 test cases passed.

Runtime : 5ms

0 0