1
# Python
2
class Solution:
3
    def maxVowels(self,s,k):
4
        ans=vowel=0
5
        for i,c in enumerate(s):
6
            # 入
7
            if c in "aeiou":
8
                vowel+=1
9
            if i < k-1:
10
                continue
11
            # 更新
12
            ans=max(ans,vowel)
13

14
            # 出
15
            if s[i-k+1] in "aeiou":
16
                voewl-=1
17
        return ans

1
// Java
2
class Solution:{
3
    public int maxVowels(String S, int k){
4
        // toCharArray(): 将一个字符串（String）转换成一个新的字符数组（char[]）
5
        char[] s=S.toCharArray();
6
        int ans =0;
7
        int vowel=0;
8
        for (int i =0;i<s.length;i++){
9
            // 入
10
            if (s[i]=='a'|| ... || s[i]=='u'){
11
                vowel++;
12
            }
13
            if (i<k-1){
14
                continue;
15
            }
16
            //   update
17
            ans=Math.max(ans,vowel);
18
            // 出
19
            char out=s[i-k+1];
20
            if (out=='a'||...||out=='u'){
21
                vowel--;
22
            }
23
        }
24
        return ans;
25
    }
26
}

1
// JavaScript
2
var maxVowels=function(s,k){
3
    let ans=0, vowel=0;
4
    for (let i=0;i<s.length;i++){
5
        // in
6
        if (s[i]==='a'||...||s[i]==='u'){
7
            vowel++;
8
        }
9
        if (i<k-1){
10
            continue;
11
        }
12
        // update
13
        ans=Math.max(ans,vowel);
14

15
        // out
16
        let out=s[i-k+1];
17
        if (out[i]==='a'||...||out[i]==='u'){
18
            vowel--;
19
        }
20
    }
21
    return ans;
22
};

1
// C++
2
class Solution{
3
    public:
4
      int maxVowels:(string s, int k){
5
            int ans=0, vowel=0;
6
            for (int i=0; i<s.size();i++){
7
                // in
8
                if (s[i]=='a'||...||s[i]=='u'){
9
                    vowel++;
10
                }
11
                int left=i-k+1;
12
                if (left<0){
13
                    continue;
14
                }
15

16
                // update
17

18
                ans=max(ans,voewl);
19

20
                // out
21

22
                char out=s[left];
23
                if (out[i]=='a'||...||out[i]=='u'){
24
                    vowel--;
25
                }
26
            }
27
            return ans;
28
        }
29
};

1
# python
2
class Solution:
3
    def searchRange(self, nums,target):
4
        # 查找当前位置的下标值
5
        start=self.lower_bound(nums,target)
6
        # nums中不存在target值
7
        if start ==len(nums) or nums[start]!=target:
8
            return [-1,-1]
9
        # 查找比当前值大一点的下标值
10
         end=self.lower_bound(nums,target+1)-1
11
        return [start,end]
12

13
    def lower_bound(self,nums,target):
14
        # 二分查找当前值在数组中的下标是哪
15
        l,r=0,len(nums)
16
        while l+1<r:
17
            mid=(l+r)//2
18
            if nums[mid]>=target:
19
                r=mid
20
            else:
21
                l=mid
22
        return l

1
//javascript
2
var searchRange=function(nums,target){
3
    const start=lowerBound(nums,target);
4
    // no exist
5
    if (start===nums.length||nums[start]!==target){
6
        return [-1,-1];
7
    }
8
    const end=lowerBound(nums,target+1)-1;
9
    return [start,end]
10

11
}
12

13
var lowerBound=function(nums,target){
14
    let l=0,r=nums.length;
15
    while (l+1<r){
16
        const mid=Math.floor((l+r)/2)
17
        if (nums[mid]>=target){
18
            r=mid;
19
        }
20
        else{
21
            l=mid;
22
        }
23
    }
24
    return l;
25
}

1
class Solution{
2
public:
3
    vector<int> searchRange(vector<int>& nums,int target){
4
        int start=lower_bound(nums,target);
5
        if (start==nums.size()||nums[start]!=target){
6
      return {-1,-1}
7
        }
8
        int end=lower_bound(nums,target+1)-1;
9
        return {start,end};
10
    }
11

12
    int lower_bound(vector<int>nums,int target){
13
        int l,r=0,nums.size();
14
        while (l+1<r){
15
            int mid=l+(r-l)/2;
16
            if (nums[mid]>=target){
17
                r=mid;
18
            }
19
            else{
20
                l=mid;
21
            }
22
        }
23
        return l;
24
    }
25

26
};

1
# Python
2
class Solution:
3
    def binarySearchMin(self,nums):
4
        def check(mid):
5
            continue
6
        l,r=..,..
7
        while l+1<r:
8
            mid=(l+r)//2
9
            if check(mid):
10
                r=mid
11
            else:
12
                l=mid
13
        return r
14

15
# 求最大值不过是把left和right替换一下
16
# 因为值比较大的贴近于right, 让left趋近right以获取最大值

1
// JavaScript
2
var Solution=function(nums){
3
  let l,r=..,..
4
    while (l+1<r){
5
        let mid=Math.floor((l+r)/2);
6
        if check(mid){
7
            r=mid;
8
        }
9
        else{
10
            l=mid;
11
        }
12
    }
13
    return r
14
};
15
var check=function(nums){
16
    // TODO
17
}

1
//c++
2
class Solution{
3
    public:
4
    int binarySearchMax(vector<int>& nums){
5
        auto check=[&](int mid){
6

7
        };
8
        int l,r=..,.;
9
        while (l+1<r){
10
            int mid=l+((r-l)/2);
11
            if (check(mid)){
12
                r=mid;
13
            }
14
            else{
15
                l=mid;
16
            }
17
        }
18
        return l;
19
    }
20
};