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CyC2018 2018-03-27 11:02:30 +08:00
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notes/剑指 offer 题解.md
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@ -2038,25 +2038,19 @@ public int FirstNotRepeatingChar(String str) {
}
```
以上的空间复杂度还不是最优的。考虑到最需要找到只出现一次的字符，那么我们只需要统计的次数信息只有 0,1,更大，那么使用两个比特位就能存储这些信息。
以上的空间复杂度还不是最优的。考虑到只需要找到只出现一次的字符，那么我们只需要统计的次数信息只有 0,1,更大，那么使用两个比特位就能存储这些信息。
```java
public int FirstNotRepeatingChar(String str) {
BitSet bs1 = new BitSet(256);
BitSet bs2 = new BitSet(256);
for (int i = 0; i < str.length(); i++) {
char c = str.charAt(i);
if (!bs1.get(c) && !bs2.get(c)) { // 0 0
bs1.set(c);
} else if (bs1.get(c) && !bs2.get(c)) { // 0 1
bs2.set(c);
}
for (char c : str.toCharArray()) {
if (!bs1.get(c) && !bs2.get(c)) bs1.set(c); // 0 0
else if (bs1.get(c) && !bs2.get(c)) bs2.set(c); // 0 1
}
for (int i = 0; i < str.length(); i++) {
char c = str.charAt(i);
if (bs1.get(c) && !bs2.get(c)) {
return i;
}
if (bs1.get(c) && !bs2.get(c)) return i;
}
return -1;
}