4949
5050### 方法一:滑动窗口
5151
52- 定义 $s$ 表示所有不满意的顾客总数,$cs$ 表示顾客总数。
52+ 根据题目描述,我们只需要统计老板不生气时的客户数量 $tot,ドル再加上一个大小为 ` minutes ` 的滑动窗口中,老板生气时的客户数量的最大值 $mx$ 即可。
53+ 54+ 我们定义一个变量 $cnt$ 来记录滑动窗口中老板生气时的客户数量,初始值为前 ` minutes ` 分钟老板生气时的客户数量。然后我们遍历数组,每次移动滑动窗口时,更新 $cnt$ 的值,同时更新 $mx$ 的值。
55+ 56+ 最后返回 $tot + mx$ 即可。
57+ 58+ 时间复杂度 $O(n),ドル其中 $n$ 为数组 ` customers ` 的长度。空间复杂度 $O(1)$。
5359
5460<!-- tabs:start -->
5561
@@ -58,36 +64,32 @@ class Solution:
5864 def maxSatisfied (
5965 self customers : List[int ], grumpy : List[int ], minutes : int
6066 ) -> int :
61- s = sum (a * b for a, b in zip (customers, grumpy))
62- cs = sum (customers)
63- t = ans = 0
64- for i, (a, b) in enumerate (zip (customers, grumpy), 1 ):
65- t += a * b
66- if (j := i - minutes) >= 0 :
67- ans = max (ans, cs - (s - t))
68- t -= customers[j] * grumpy[j]
69- return ans
67+ mx = cnt = sum (c * g for c, g in zip (customers[:minutes], grumpy))
68+ for i in range (minutes, len (customers)):
69+ cnt += customers[i] * grumpy[i]
70+ cnt -= customers[i - minutes] * grumpy[i - minutes]
71+ mx = max (mx, cnt)
72+ return sum (c * (g ^ 1 ) for c, g in zip (customers, grumpy)) + mx
7073```
7174
7275``` java
7376class Solution {
7477 public int maxSatisfied (int [] customers , int [] grumpy , int minutes ) {
75- int s = 0 , cs = 0 ;
76- int n = customers . length ;
77- for (int i = 0 ; i < n ; ++ i) {
78- s += customers[i] * grumpy[i];
79- cs += customers[i];
78+ int cnt = 0 ;
79+ int tot = 0 ;
80+ for (int i = 0 ; i < minutes ; ++ i) {
81+ cnt += customers[i] * grumpy[i];
82+ tot += customers[i]* (grumpy[i] ^ 1 ) ;
8083 }
81- int t = 0 , ans = 0 ;
82- for (int i = 0 ; i < n; ++ i) {
83- t += customers[i] * grumpy[i];
84- int j = i - minutes + 1 ;
85- if (j >= 0 ) {
86- ans = Math . max(ans, cs - (s - t));
87- t -= customers[j] * grumpy[j];
88- }
84+ int mx = cnt;
85+ int n = customers. length;
86+ for (int i = minutes; i < n; ++ i) {
87+ cnt += customers[i] * grumpy[i];
88+ cnt -= customers[i - minutes] * grumpy[i - minutes];
89+ mx = Math . max(mx, cnt);
90+ tot += customers[i] * (grumpy[i] ^ 1 );
8991 }
90- return ans ;
92+ return tot + mx ;
9193 }
9294}
9395```
@@ -96,76 +98,80 @@ class Solution {
9698class Solution {
9799public:
98100 int maxSatisfied(vector<int >& customers, vector<int >& grumpy, int minutes) {
99- int s = 0, cs = 0;
100- int n = customers.size() ;
101- for (int i = 0; i < n ; ++i) {
102- s += customers[ i] * grumpy[ i] ;
103- cs += customers[ i] ;
101+ int cnt = 0;
102+ int tot = 0 ;
103+ for (int i = 0; i < minutes ; ++i) {
104+ cnt += customers[ i] * grumpy[ i] ;
105+ tot += customers[ i] * (grumpy [ i ] ^ 1) ;
104106 }
105- int t = 0, ans = 0;
106- for (int i = 0; i < n; ++i) {
107- t += customers[ i] * grumpy[ i] ;
108- int j = i - minutes + 1;
109- if (j >= 0) {
110- ans = max(ans, cs - (s - t));
111- t -= customers[ j] * grumpy[ j] ;
112- }
107+ int mx = cnt;
108+ int n = customers.size();
109+ for (int i = minutes; i < n; ++i) {
110+ cnt += customers[ i] * grumpy[ i] ;
111+ cnt -= customers[ i - minutes] * grumpy[ i - minutes] ;
112+ mx = max(mx, cnt);
113+ tot += customers[ i] * (grumpy[ i] ^ 1);
113114 }
114- return ans ;
115+ return tot + mx ;
115116 }
116117};
117118```
118119
119120```go
120121func maxSatisfied(customers []int, grumpy []int, minutes int) int {
121- s, cs := 0, 0
122- for i, c := range customers {
123- s += c * grumpy[i]
124- cs += c
122+ var cnt, tot int
123+ for i, c := range customers[:minutes] {
124+ cnt += c * grumpy[i]
125+ tot += c * (grumpy[i] ^ 1)
125126 }
126- t, ans := 0, 0
127- for i, c := range customers {
128- t += c * grumpy[i]
129- j := i - minutes + 1
130- if j >= 0 {
131- ans = max(ans, cs-(s-t))
132- t -= customers[j] * grumpy[j]
133- }
127+ mx := cnt
128+ for i := minutes; i < len(customers); i++ {
129+ cnt += customers[i] * grumpy[i]
130+ cnt -= customers[i-minutes] * grumpy[i-minutes]
131+ mx = max(mx, cnt)
132+ tot += customers[i] * (grumpy[i] ^ 1)
134133 }
135- return ans
134+ return tot + mx
135+ }
136+ ```
137+ 138+ ``` ts
139+ function maxSatisfied(customers : number [], grumpy : number [], minutes : number ): number {
140+ let [cnt, tot] = [0 , 0 ];
141+ for (let i = 0 ; i < minutes ; ++ i ) {
142+ cnt += customers [i ] * grumpy [i ];
143+ tot += customers [i ] * (grumpy [i ] ^ 1 );
144+ }
145+ let mx = cnt ;
146+ for (let i = minutes ; i < customers .length ; ++ i ) {
147+ cnt += customers [i ] * grumpy [i ];
148+ cnt -= customers [i - minutes ] * grumpy [i - minutes ];
149+ mx = Math .max (mx , cnt );
150+ tot += customers [i ] * (grumpy [i ] ^ 1 );
151+ }
152+ return tot + mx ;
136153}
137154```
138155
139156``` rust
140157impl Solution {
141158 pub fn max_satisfied (customers : Vec <i32 >, grumpy : Vec <i32 >, minutes : i32 ) -> i32 {
142- let k = minutes as usize ;
143- let n = customers . len ();
144- 145- let mut sum = 0 ;
146- for i in 0 .. k {
147- if grumpy [i ] == 1 {
148- sum += customers [i ];
149- }
150- }
151- let mut max = sum ;
152- for i in k .. n {
153- if grumpy [i - k ] == 1 {
154- sum -= customers [i - k ];
155- }
156- if grumpy [i ] == 1 {
157- sum += customers [i ];
158- }
159- max = max . max (sum );
159+ let mut cnt = 0 ;
160+ let mut tot = 0 ;
161+ let minutes = minutes as usize ;
162+ for i in 0 .. minutes {
163+ cnt += customers [i ] * grumpy [i ];
164+ tot += customers [i ] * (1 - grumpy [i ]);
160165 }
161- 162- sum = 0 ;
163- for i in 0 .. n {
164- if grumpy [i ] == 0 {
165- sum += customers [i ];
166- }
166+ let mut mx = cnt ;
167+ let n = customers . len ();
168+ for i in minutes .. n {
169+ cnt += customers [i ] * grumpy [i ];
170+ cnt -= customers [i - minutes ] * grumpy [i - minutes ];
171+ mx = mx . max (cnt );
172+ tot += customers [i ] * (1 - grumpy [i ]);
167173 }
168- sum + max
174+ tot + mx
169175 }
170176}
171177```
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