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剑指Offer03. 数组中重复的数字【简单】
2022-07-03 11:50:00 【伍六琪】
剑指 Offer 03. 数组中重复的数字
题目描述:
找出数组中重复的数字。
在一个长度为 n 的数组 nums 里的所有数字都在 0~n-1 的范围内。数组中某些数字是重复的,但不知道有几个数字重复了,也不知道每个数字重复了几次。请找出数组中任意一个重复的数字。
示例 1:
> 输入: [2, 3, 1, 0, 2, 5, 3]
> 输出:2 或 3
限制:
2 <= n <= 100000
JAVA代码
普通方法
两层遍历挨个寻找第一个重复的数字。
class Solution {
public int findRepeatNumber(int[] nums) {
for(int i = 0;i<nums.length-1;i++){
for(int j = i+1;j<nums.length;j++){
if(nums[i]==nums[j]){
return nums[i];
}
}
}
return -1;
}
}
Map方法
使用map存储nums数据,以nums的值作为map的key,当遇到相同的key值时返回该数值。
class Solution {
public int findRepeatNumber(int[] nums) {
Map<Integer,Integer> map = new HashMap<Integer,Integer>();
for(int i = 0;i<nums.length;i++){
if(map.containsKey(nums[i])){
return nums[i];
}
map.put(nums[i],i);
}
return -1;
}
}
官方方法
使用Set,仅存储一个值,比Map方法存储两个值更加简便
拓展:Set无序的不可重复的。add()方法返回false,则说明存在重复的值。
class Solution {
public int findRepeatNumber(int[] nums) {
Set<Integer> set = new HashSet<Integer>();
int result = -1;
for(int num:nums){
if(!set.add(num)){
result = num;
break;
}
}
return result;
}
}
“这里没有涉及到其他复杂的关于位置的操作,所以仅仅存储值就足够了。”
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