1.题目:
2.解析:
方法一:用哈希表:记录存在的数字,找到哈希表为空的数字输出Set<Integer> set = new HashSet<>();for(int x : records) set.add(x);for(int i = 0; i < set.size(); i++){if(!set.contains(i)) return i;}return set.size();
方法二:位运算异或
int ret = 0;for (int i = 0; i < records.length; i++) {ret ^= records[i];ret ^= i + 1;}return ret;
方法三:直接遍历
int ret = 0;for(int i = 0; i < records.length; i++){if(records[i] != i) return i;}return records.length;
方法四:二分查找:
注意:特殊情况整个数组元素及对应下标都一样时
int left = 0,right = records.length-1;while(left < right){int mid = left + (right - left) / 2;if(records[mid] == mid) left = mid+1;else right = mid;}//特殊情况:整个数组元素及对应下标都一样时return records[left] == left ? left+1 : left;