LeetCode & Q268-Missing Number-Easy
Array
Math
Bit Manipulation
Description:
Given an array containing n distinct numbers taken from
0, 1, 2, ..., n
, find the one that is missing from the array.For example,
Given nums =[0, 1, 3]
return2
.
这题真是一言难尽....刚开始不太理解,数字要从0开始往后记,我本来写的Binary Search有问题...参考了Discuss里重写了一遍
看到Discuss里很多用XOR写的,以前没有涉及过这个领域,感觉好神奇...
my Solution:
public class Solution {
public int missingNumber(int[] nums) {
Arrays.sort(nums);
int left = 0, right = nums.length, mid= (left + right)/2;
while(left < right){
mid = (left + right) / 2;
if(nums[mid] > mid) right = mid;
else left = mid + 1;
}
return left;
}
}
XOR Solution:
// a^b^b = a, nums[index] = index
public int missingNumber(int[] nums) {
int xor = 0, i = 0;
for (i = 0; i < nums.length; i++) {
xor = xor ^ i ^ nums[i];
}
return xor ^ i;
}
看到最快的是用数学稍微转换下思路的方法,数学好的人真厉害...其实也不是特别厉害的数学....可是我怎么就没想起来呢...
public class Solution {
public int missingNumber(int[] nums) {
int n = nums.length;
int sum = (n*(n+1)) /2;
int numSum =0;
for(int i =0;i<nums.length;i++){
numSum += nums[i];
}
int missingNumber = sum - numSum;
return missingNumber;
}
}