Poison

268. Missing Number

发表于
Math
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class Solution {
    public int missingNumber(int[] nums) {
        int exceptedSum = nums.length * (nums.length + 1) / 2;
        int sum = 0;
        for (int num : nums) {
            sum += num;
        }
        return exceptedSum - sum;
    }
}
Bit
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class Solution {
    public int missingNumber(int[] nums) {
        int res = 0;
        for (int i = 0; i < nums.length; i++) {
            res ^= nums[i];
            res ^= i; // 此处处理 [0, n - 1]
        }
        res ^= nums.length; // 未丢失数字时整个范围为 [0, n], 此时我们对 for 循环未异或的最后一个 n 进行异或处理
        return res;
    }
}
Reference

268. Missing Number