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Commit d8ec3b6

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feat: add solutions to lcp problem: No.68
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‎lcp/LCP 68. 美观的花束/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:双指针**
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我们用双指针 $j$ 和 $i$ 分别指向当前窗口的左右端点,用数组或哈希表 $d$ 记录当前窗口内的元素以及出现的次数。
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遍历数组 $flowers,ドル每一次我们将 $flowers[i]$ 加入到窗口中,即 $d[flowers[i]]++,ドル然后判断 $d[flowers[i]]$ 是否大于 $cnt,ドル如果大于 $cnt,ドル则我们需要将 $flowers[j]$ 从窗口中移除,即 $d[flowers[j]]--,ドル并将 $j$ 右移,直到 $d[flowers[i]] \leq cnt$。此时窗口内的元素都不超过 $cnt$ 个,因此我们可以将 $i - j + 1$ 加到答案中。
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最后返回答案即可。
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时间复杂度 $O(n),ドル空间复杂度 $O(m)$。其中 $n$ 和 $m$ 分别为数组 $flowers$ 的长度以及数组 $flowers$ 中的最大值。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def beautifulBouquet(self, flowers: List[int], cnt: int) -> int:
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mod = 10**9 + 7
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d = Counter()
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ans = j = 0
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for i, x in enumerate(flowers):
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d[x] += 1
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while d[x] > cnt:
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d[flowers[j]] -= 1
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j += 1
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ans = (ans + i - j + 1) % mod
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int beautifulBouquet(int[] flowers, int cnt) {
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int mx = 0;
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for (int x : flowers) {
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mx = Math.max(mx, x);
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}
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int[] d = new int[mx + 1];
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long ans = 0;
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final int mod = (int) 1e9 + 7;
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for (int i = 0, j = 0; i < flowers.length; ++i) {
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++d[flowers[i]];
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while (d[flowers[i]] > cnt) {
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--d[flowers[j++]];
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}
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ans = (ans + i - j + 1) % mod;
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}
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return (int) ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int beautifulBouquet(vector<int>& flowers, int cnt) {
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int mx = *max_element(flowers.begin(), flowers.end());
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int d[mx + 1];
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memset(d, 0, sizeof(d));
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long long ans = 0;
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const int mod = 1e9 + 7;
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for (int i = 0, j = 0; i < flowers.size(); ++i) {
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++d[flowers[i]];
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while (d[flowers[i]] > cnt) {
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--d[flowers[j++]];
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}
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ans = (ans + i - j + 1) % mod;
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func beautifulBouquet(flowers []int, cnt int) (ans int) {
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mx := 0
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for _, x := range flowers {
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mx = max(mx, x)
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}
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d := make([]int, mx+1)
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j := 0
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const mod = 1e9 + 7
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for i, x := range flowers {
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d[x]++
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for d[x] > cnt {
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d[flowers[j]]--
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j++
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}
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ans = (ans + i - j + 1) % mod
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**
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class Solution {
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public:
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int beautifulBouquet(vector<int>& flowers, int cnt) {
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int mx = *max_element(flowers.begin(), flowers.end());
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int d[mx + 1];
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memset(d, 0, sizeof(d));
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long long ans = 0;
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const int mod = 1e9 + 7;
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for (int i = 0, j = 0; i < flowers.size(); ++i) {
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++d[flowers[i]];
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while (d[flowers[i]] > cnt) {
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--d[flowers[j++]];
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}
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ans = (ans + i - j + 1) % mod;
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}
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return ans;
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}
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};
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func beautifulBouquet(flowers []int, cnt int) (ans int) {
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mx := 0
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for _, x := range flowers {
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mx = max(mx, x)
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}
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d := make([]int, mx+1)
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j := 0
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const mod = 1e9 + 7
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for i, x := range flowers {
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d[x]++
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for d[x] > cnt {
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d[flowers[j]]--
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j++
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}
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ans = (ans + i - j + 1) % mod
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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class Solution {
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public int beautifulBouquet(int[] flowers, int cnt) {
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int mx = 0;
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for (int x : flowers) {
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mx = Math.max(mx, x);
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}
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int[] d = new int[mx + 1];
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long ans = 0;
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final int mod = (int) 1e9 + 7;
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for (int i = 0, j = 0; i < flowers.length; ++i) {
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++d[flowers[i]];
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while (d[flowers[i]] > cnt) {
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--d[flowers[j++]];
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}
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ans = (ans + i - j + 1) % mod;
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}
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return (int) ans;
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}
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}
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class Solution:
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def beautifulBouquet(self, flowers: List[int], cnt: int) -> int:
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mod = 10**9 + 7
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d = Counter()
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ans = j = 0
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for i, x in enumerate(flowers):
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d[x] += 1
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while d[x] > cnt:
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d[flowers[j]] -= 1
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j += 1
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ans = (ans + i - j + 1) % mod
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return ans

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