[LeetCode] 8. String to Integer (atoi) java
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/**8. String to Integer (atoi) * @param str * @returnint 字符串->数字 */ public int myAtoi(String str) { if (str == null || str.length() == 0) return 0; boolean negetive = true; boolean start = false; double ret = 0; for (int i=0, len=str.length(); i<len; i++) { char ch = str.charAt(i); if (!start && ch == '-') { start = true; negetive = false; } else if (!start && ch == '+') { start = true; negetive = true; }else if (!start && ch == ' '){ continue; }else if (!(ch>='0' && ch<='9')) { break; }else { start = true; ret = ret*10 + (ch-'0'); } } ret = negetive? ret: -ret; if (negetive && ret > Integer.MAX_VALUE) { return Integer.MAX_VALUE; }else if (!negetive && ret < Integer.MIN_VALUE) { return Integer.MIN_VALUE; } return (int)ret; }
考虑因素:+-,空额,溢出,不考虑E
” -+00123 23”
解决办法:使用double双精度存储结果,最后和最大值最小值比较
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