Java C++题解leetcode消失的两个数字实例

class Solution {
    public int[] missingTwo(int[] nums) {
        int len = nums.length + 2;
        int tot = len * (1 + len) / 2;
        for (int x : nums)
            tot -= x;
        int sum = tot, m = tot / 2;
        tot = m * (1 + m) / 2;
        for (int x : nums) {
            if (x <= m) // m向下取整，可能相等
                tot -= x;
        }
        return new int[]{tot, sum - tot};
    }
}

class Solution {
public:
    vector<int> missingTwo(vector<int>& nums) {
        int len = nums.size() + 2;
        int tot = len * (1 + len) / 2;
        for (int x : nums)
            tot -= x;
        int sum = tot, m = tot / 2;
        tot = m * (1 + m) / 2;
        for (int x : nums) {
            if (x <= m) // m向下取整，可能相等
                tot -= x;
        }
        return {tot, sum - tot};
    }
};

impl Solution {
    pub fn missing_two(nums: Vec<i32>) -> Vec<i32> {
        let len = nums.len() as i32 + 2;
        let mut sum : i32 = nums.iter().sum();
        sum = len * (1 + len) / 2 - sum;
        let m = sum / 2; // m向下取整，可能相等
        let mut lsum : i32 = nums.iter().filter(|&x| x <= &m).sum();
        lsum = m * (1 + m) / 2 - lsum;
        vec![lsum, sum - lsum]
    }
}