java中字符串转整数及MyAtoi方法的实现

"  010"
"  +004500"
" -001+2a42"
"  +0 123"
"-2147483648"
"2147483648"
"  - 321"
"   -11919730356x"
"9223372036854775809"

10
4500
-1
0
-2147483648
2147483647
0
-2147483648
2147483647

public static int myAtoi(String str) { 
    if(str==null||str.length()==0) 
      return 0; 
    char[] array = str.toCharArray(); 
    long result = 0; // 要返回的结果result 
    int count = 0; // 记录‘+'或者‘-'出现的次数 
    int num = 0;  // 判断空格出现的位置 
    int flag = 1; // 正数还是负数 
    for (int i = 0; i < array.length; i++) { 
      Character c = array[i]; 
      if(c>='0'&&c<='9'){ 
        result = result*10+c-'0'; 
        // 判断是否溢出 
        if(flag==1&&result>Integer.MAX_VALUE){ 
          return Integer.MAX_VALUE; 
        }else if(flag==-1&&-result<Integer.MIN_VALUE) 
          return Integer.MIN_VALUE; 
        num++; 
      }else if(c==' '&&num==0&&count==0) 
        continue; 
      else if(c=='+'&&count==0){ 
        count = 1; 
      } 
      else if(c=='-'&&count==0){ 
        flag = -1; 
        count = 1; 
      } 
      else{ 
        return (int) (flag*result); 
         
      } 
    } 
    return (int) (flag*result); 
  } 

for (int i = 0; i < array.length; i++) { 
      Character c = array[i]; 
      if(Character.isDigit(c)){ 
        result = result*10+Character.getNumericValue(c); 
        if(flag==1&&result>Integer.MAX_VALUE){ 
          return Integer.MAX_VALUE; 
        }else if(flag==-1&&-result<Integer.MIN_VALUE) 
          return Integer.MIN_VALUE; 
        num++; 
      }else if(Character.isSpaceChar(c)&&num==0&&count==0) 
        continue; 
      else if(c=='+'&&count==0){ 
        count = 1; 
      } 
      else if(c=='-'&&count==0){ 
        flag = -1; 
        count = 1; 
      } 
      else{ 
        return (int) (flag*result); 
         
      } 
    }