Solution 1: No '1' After '0'

Notice that the string $s$ does not contain leading zeros, which means $s$ starts with '1'.

If the string $s$ contains the substring "01", then $s$ must be a string like "1...01...", in which case $s$ has at least two consecutive '1' segments, which does not satisfy the problem condition, so we return false.

If the string $s$ does not contain the substring "01", then $s$ can only be a string like "1..1000...", in which case $s$ has only one consecutive '1' segment, which satisfies the problem condition, so we return true.

Therefore, we only need to judge whether the string $s$ contains the substring "01".

The time complexity is $O (n)$ , where $n$ is the length of the string $s$ . The space complexity is $O (1)$ .

Python3

class Solution:
    def checkOnesSegment(self, s: str) -> bool:
        return '01' not in s

Java

class Solution {
    public boolean checkOnesSegment(String s) {
        return !s.contains("01");
    }
}

C++

class Solution {
public:
    bool checkOnesSegment(string s) {
        return s.find("01") == -1;
    }
};

Go

func checkOnesSegment(s string) bool {
    return !strings.Contains(s, "01")
}

TypeScript

function checkOnesSegment(s: string): boolean {
    let pre = s[0];
    for (const c of s) {
        if (pre !== c && c === '1') {
            return false;
        }
        pre = c;
    }
    return true;
}

Rust

impl Solution {
    pub fn check_ones_segment(s: String) -> bool {
        !s.contains("01")
    }
}

Problem