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LeetCode1052.java
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LeetCode1052.java
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/**
* @ProjectName: LeetCode
* @Author: XinyuLiu
* @Description:
*/
public class LeetCode1052 {
public int maxSatisfied(int[] customers, int[] grumpy, int X) {
int total = 0;
for (int i = 0 ; i < customers.length ; i++){
if(grumpy[i] == 0){total += customers[i];}
}
int max = 0;
for (int i = 0; i <customers.length-X+1 ; i++) {
int a=0;
for (int j = i;j<i+X;j++) {
if(grumpy[j] == 1){a += customers[j];}
}
if(a>max){max = a;}
}
return total+max;
}
public int maxSatisfied2(int[] customers, int[] grumpy, int X) {
//前X位的计数
int count = 0;
for (int n=0;n<X;n++){
count = count + customers[n]*grumpy[n];
}
int max = count;
//寻找最大生气区间
for (int i=1;i<=customers.length-X;i++){
count = count - customers[i-1]*grumpy[i-1] + customers[i+X-1]*grumpy[i+X-1];//很巧妙 省时
if (count>max){
max = count;
}
}
int res = 0;
//优化 使用「秘密技巧」能得到的最终的顾客数 = 所有不生气时间内的顾客总数 + 在窗口 X 内使用「秘密技巧」挽留住的原本因为生气而被赶走顾客数。
for (int i=0;i<customers.length;i++){
if (grumpy[i]==0){
res = res + customers[i];
}
}
res = res + max;
return res;
}
}