题目描述
给定一个只包含整数的有序数组，每个元素都会出现两次，唯有一个数只会出现一次，找出这个数。
示例:
输入: [1,1,2,3,3,4,4,8,8]
输出: 2
示例2:
输入: [3,3,7,7,10,11,11]
输出: 10
题目思路
暴力法
直接查找元素
class Solution {
public int singleNonDuplicate(int[] nums) {
for (int i = 0; i < nums.length - 1; i += 2) {
if(nums[i]!=nums[i+1]){
return nums[i];
}
}
return nums[nums.length-1];
}
}
哈希表查找
public int singleNonDuplicate(int[] nums) {
HashMap<Integer, Integer> map = new HashMap<>();
for (int num : nums) {
map.put(num, map.getOrDefault(num, 0) + 1);
}
int key = 0;
for (Integer integer : map.keySet()) {
if (map.get(integer).equals(1)){
key = integer;
}
}
return key;
}
二分查找
public int singleNonDuplicate(int[] nums) {
int lo = 0;
int hi = nums.length - 1;
while (lo < hi) {
int mid = lo + (hi - lo) / 2;
boolean halvesAreEven = (hi - mid) % 2 == 0;
if (nums[mid + 1] == nums[mid]) {
if (halvesAreEven) {
lo = mid + 2;
} else {
hi = mid - 1;
}
} else if (nums[mid - 1] == nums[mid]) {
if (halvesAreEven) {
hi = mid - 2;
} else {
lo = mid + 1;
}
} else {
return nums[mid];
}
}
return nums[lo];
}