풀이: 릿코드.1784.Check if Binary String Has at Most One Segment of Ones

?: 순회를 이용해 풀이

Author: changicho

LeetCode/easy/1784. Check if Binary String Has at Most One Segment of Ones/README.md

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+# 1784. Check if Binary String Has at Most One Segment of Ones
+
+[링크](https://leetcode.com/problems/check-if-binary-string-has-at-most-one-segment-of-ones/description/)
+
+| 난이도 |
+| :----: |
+|  Easy  |
+
+## 설계
+
+### 시간 복잡도
+
+문자열의 길이를 N이라 하자.
+
+순회하며 1이 나오는 구간의 수를 셀 경우 O(N)의 시간 복잡도를 사용한다.
+
+### 공간 복잡도
+
+순회에 O(1)의 공간 복잡도를 사용한다.
+
+### 순회
+
+| 내 코드 (ms) | 시간 복잡도 | 공간 복잡도 |
+| :----------: | :---------: | :---------: |
+|      0       |    O(N)     |    O(1)     |
+
+```cpp
+bool checkOnesSegment(string s) {
+  int size = s.size();
+  int count = 0;
+
+  char before = '0';
+  for (int i = 0; i < size; i++) {
+    if (s[i] == '0' && before == '1') {
+      count++;
+
+      if (count > 1) return false;
+    }
+    before = s[i];
+  }
+  if (before == '1') count++;
+
+  return count <= 1;
+}
+```
+
+## 고생한 점

LeetCode/easy/1784. Check if Binary String Has at Most One Segment of Ones/code.cpp

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+#include <algorithm>
+#include <climits>
+#include <cmath>
+#include <cstring>
+#include <functional>
+#include <iostream>
+#include <map>
+#include <numeric>
+#include <queue>
+#include <set>
+#include <stack>
+#include <string>
+#include <unordered_map>
+#include <unordered_set>
+#include <vector>
+
+using namespace std;
+
+// one pass
+// time : O(N)
+// space : O(1)
+class Solution {
+ public:
+  bool checkOnesSegment(string s) {
+    int size = s.size();
+    int count = 0;
+
+    char before = '0';
+    for (int i = 0; i < size; i++) {
+      if (s[i] == '0' && before == '1') {
+        count++;
+
+        if (count > 1) return false;
+      }
+      before = s[i];
+    }
+    if (before == '1') count++;
+
+    return count <= 1;
+  }
+};

LeetCode/easy/1784. Check if Binary String Has at Most One Segment of Ones/input.txt

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+Input: s = "1001"
+Output: false
+
+===
+
+Input: s = "110"
+Output: true