2996. 大于等于顺序前缀和的最小缺失整数

class Solution:
    def missingInteger(self, nums: List[int]) -> int:
        s, j = nums[0], 1
        while j < len(nums) and nums[j] == nums[j - 1] + 1:
            s += nums[j]
            j += 1
        vis = set(nums)
        for x in count(s):
            if x not in vis:
                return x

class Solution {
    public int missingInteger(int[] nums) {
        int s = nums[0];
        for (int j = 1; j < nums.length && nums[j] == nums[j - 1] + 1; ++j) {
            s += nums[j];
        }
        boolean[] vis = new boolean[51];
        for (int x : nums) {
            vis[x] = true;
        }
        for (int x = s;; ++x) {
            if (x >= vis.length || !vis[x]) {
                return x;
            }
        }
    }
}

class Solution {
public:
    int missingInteger(vector<int>& nums) {
        int s = nums[0];
        for (int j = 1; j < nums.size() && nums[j] == nums[j - 1] + 1; ++j) {
            s += nums[j];
        }
        bitset<51> vis;
        for (int x : nums) {
            vis[x] = 1;
        }
        for (int x = s;; ++x) {
            if (x >= 51 || !vis[x]) {
                return x;
            }
        }
    }
};

func missingInteger(nums []int) int {
	s := nums[0]
	for j := 1; j < len(nums) && nums[j] == nums[j-1]+1; j++ {
		s += nums[j]
	}
	vis := [51]bool{}
	for _, x := range nums {
		vis[x] = true
	}
	for x := s; ; x++ {
		if x >= len(vis) || !vis[x] {
			return x
		}
	}
}

function missingInteger(nums: number[]): number {
    let s = nums[0];
    for (let j = 1; j < nums.length && nums[j] === nums[j - 1] + 1; ++j) {
        s += nums[j];
    }
    const vis: Set<number> = new Set(nums);
    for (let x = s; ; ++x) {
        if (!vis.has(x)) {
            return x;
        }
    }
}