feat: add solutions to lc problems: No.1869~1872 #1980

Original file line number	Diff line number	Diff line change
Expand Up		@@ -62,9 +62,11 @@

		<!-- 这里可写通用的实现逻辑 -->

	直接遍历字符串,获取"0 子串"和"1 子串"的最大长度 `len0`、`len1`。
	方法一:两次遍历

	遍历结束后,若 `len1 > len0`,返回 true,否则返回 false。
	我们设计一个函数 $f(x),ドル表示字符串 $s$ 中由 $x$ 组成的最长连续子字符串的长度。如果 $f(1) \gt f(0),ドル那么返回 `true`,否则返回 `false`。

	时间复杂度 $O(n),ドル其中 $n$ 是字符串 $s$ 的长度。空间复杂度 $O(1)$。

		<!-- tabs:start -->

Expand All		@@ -75,18 +77,17 @@
		```python
		class Solution:
		def checkZeroOnes(self, s: str) -> bool:
	n0 = n1 = 0
	t0 = t1 = 0
	for c in s:
	if c == '0':
	t0 += 1
	t1 = 0
	else:
	t0 = 0
	t1 += 1
	n0 = max(n0, t0)
	n1 = max(n1, t1)
	return n1 > n0
	def f(x: str) -> int:
	cnt = mx = 0
	for c in s:
	if c == x:
	cnt += 1
	mx = max(mx, cnt)
	else:
	cnt = 0
	return mx

	return f("1") > f("0")
		```

		### Java
Expand All		@@ -96,71 +97,41 @@ class Solution:
		```java
		class Solution {
		public boolean checkZeroOnes(String s) {
	int n0 = 0, n1 = 0;
	int t0 = 0, t1 = 0;
	return f(s, '1') > f(s, '0');
	}

	private int f(String s, char x) {
	int cnt = 0, mx = 0;
		for (int i = 0; i < s.length(); ++i) {
	if (s.charAt(i) == '0') {
	++t0;
	t1 = 0;
	if (s.charAt(i) == x) {
	mx = Math.max(mx, ++cnt);
		} else {
	++t1;
	t0 = 0;
	cnt = 0;
		}
	n0 = Math.max(n0, t0);
	n1 = Math.max(n1, t1);
		}
	return n1 > n0;
	return mx;
		}
		}
		```

	### JavaScript

	```js
	/**
	* @param {string} s
	* @return {boolean}
	*/
	var checkZeroOnes = function (s) {
	let max0 = 0,
	max1 = 0;
	let t0 = 0,
	t1 = 0;
	for (let char of s) {
	if (char == '0') {
	t0++;
	t1 = 0;
	} else {
	t1++;
	t0 = 0;
	}
	max0 = Math.max(max0, t0);
	max1 = Math.max(max1, t1);
	}
	return max1 > max0;
	};
	```

		### C++

		```cpp
		class Solution {
		public:
		bool checkZeroOnes(string s) {
	int n0 = 0, n1 = 0;
	int t0 = 0, t1 = 0;
	for (auto c : s) {
	if (c == '0') {
	++t0;
	t1 = 0;
	} else {
	++t1;
	t0 = 0;
	auto f = [&](char x) {
	int cnt = 0, mx = 0;
	for (char& c : s) {
	if (c == x) {
	mx = max(mx, ++cnt);
	} else {
	cnt = 0;
	}
		}
	n0 = max(n0, t0);
	n1 = max(n1, t1);
	}
	return n1 > n0;
	return mx;
	};
	return f('1') > f('0');
		}
		};
		```
Expand All		@@ -169,23 +140,64 @@ public:

		```go
		func checkZeroOnes(s string) bool {
	n0, n1 := 0, 0
	t0, t1 := 0, 0
	for _, c := range s {
	if c == '0' {
	t0++
	t1 = 0
	} else {
	t1++
	t0 = 0
	f := func(x rune) int {
	cnt, mx := 0, 0
	for _, c := range s {
	if c == x {
	cnt++
	mx = max(mx, cnt)
	} else {
	cnt = 0
	}
		}
	n0 = max(n0, t0)
	n1 = max(n1, t1)
	return mx
		}
	return n1 > n0
	return f('1') > f('0')
		}
		```

	### TypeScript

	```ts
	function checkZeroOnes(s: string): boolean {
	const f = (x: string): number => {
	let [mx, cnt] = [0, 0];
	for (const c of s) {
	if (c === x) {
	mx = Math.max(mx, ++cnt);
	} else {
	cnt = 0;
	}
	}
	return mx;
	};
	return f('1') > f('0');
	}
	```

	### JavaScript

	```js
	/**
	* @param {string} s
	* @return {boolean}
	*/
	var checkZeroOnes = function (s) {
	const f = x => {
	let [mx, cnt] = [0, 0];
	for (const c of s) {
	if (c === x) {
	mx = Math.max(mx, ++cnt);
	} else {
	cnt = 0;
	}
	}
	return mx;
	};
	return f('1') > f('0');
	};
	```

		### ...

		```
Expand Down

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