我有一个日志项数组,已按时间戳(自 1970 年以来的毫秒数)排序。现在我想按特定时间范围过滤它们,所以我想到了二分搜索,但是这个变体与我之前知道的所有变体不同,因为我需要在一个范围内找到一个范围。请注意,值边缘可能没有或有多个项目。
我想出了这个来减少一个范围要求,但仍然不知道如何到达第一个/最后一个边缘项目:
filterByTime(min: number, max: number): LogItem[] {
const items = this.items;
const len = items.length;
if (min === undefined && max === undefined) {
return items.slice(0, len - 1);
}
min = min || Number.MIN_VALUE;
max = max || Number.MAX_VALUE;
const fn = (a: LogItem) => {
if (a.time < min) {
return -1;
} else if (a.time > max) {
return 1;
} else {
return 0;
}
};
const lowerBound = this.binarySearchBound(fn, 0, len, true);
if (lowerBound == -1) { return []; }
const upperBound = this.binarySearchBound(fn, 0, len, false);
return items.slice(lowerBound, upperBound + 1);
}
binarySearchBound(compareFn: (a: LogItem) => number, left: number, right: number, isLowerBound: boolean): number {
const items = this.items;
if (items.length == 0) {
return -1;
}
if (isLowerBound && compareFn(items[0]) == 0) {
return 0;
} else if (!isLowerBound && compareFn(items[items.length - 1]) == 0) {
return items.length - 1;
}
while (left <= right) {
const mid = (left + right) / 2;
const item = this.items[mid];
const compare = compareFn(item);
if (compare < 0) {
left = mid + 1;
} else if (compare > 0) {
right = mid - 1;
} else {
// What to do now?
}
}
return -1;
}
最坏的情况,我可以从边缘进行线性搜索,因为我可以假设边缘没有太多项目,但肯定有一个我没有想到的更好的方法,但我可能不得不迭代如果 mid 位于结果集的中间,则遍历整个结果集。 编辑添加注释:
min
或
max
可能是 undefined
(也可能是两者,在这种情况下,我可以只设置 if
并返回整个数组的副本)。如果是 MIN_VALUE
,最好用 MAX_VALUE
和 undefined
代替,还是有更好的方法来处理这种情况?
isLowerBound
布尔值不会妨碍执行时间。
upperBound
表示属于该范围的潜在最后一个索引之后的
next索引。这与参数与
slice
等本机函数的工作方式相对应。
-1
作为特殊索引。如果编码得当,无论如何,两个二分搜索都会出现一个空范围,并给出一个空数组作为结果
min
或
max
值。
MIN_VALUE
和
MAX_VALUE
作为默认值,而不测试边界情况。如果边界情况经常发生,可能值得包括这些检查,但通常要注意,这些检查将为每个过滤器执行,这可能会影响平均执行时间。
function filterByTime(min=Number.MIN_VALUE, max=Number.MAX_VALUE) {
const fn = (a, b) => a - b; // simplified (should be a.time - b.time)
const lowerBound = this.binarySearchLowerBound(fn, 0, this.items.length, min);
const upperBound = this.binarySearchUpperBound(fn, lowerBound, this.items.length, max);
return this.items.slice(lowerBound, upperBound);
}
function binarySearchLowerBound(compareFn, left, right, target) {
while (left < right) {
const mid = (left + right) >> 1;
if (compareFn(this.items[mid], target) < 0) {
left = mid + 1;
} else { // Also when equal...
right = mid;
}
}
return left;
}
function binarySearchUpperBound(compareFn, left, right, target) {
while (left < right) {
const mid = (left + right) >> 1;
if (compareFn(this.items[mid], target) <= 0) { // Also when equal...
left = mid + 1;
} else {
right = mid;
}
}
return left;
}
// Demo with simplified data (instead of objects with time property)
this.items = [1, 2, 2, 2, 3, 4, 4, 5, 5, 5, 6, 7, 8, 8];
console.log(this.filterByTime(2, 4));
console.log(this.filterByTime(4, 5));
first
和 last
代码合并到一个函数中:
filterByTime(items: LogItem[], min: number, max: number): LogItem[] {
const len = items.length;
if (len == 0) {
return [];
}
if (min === undefined && max === undefined) {
return items.slice(0, len - 1);
}
min = min || Number.MIN_VALUE;
max = max || Number.MAX_VALUE;
const fn = (a: LogItem) => {
if (a.time < min) {
return -1;
} else if (a.time > max) {
return 1;
} else {
return 0;
}
};
const lowerBound = this.binarySearchBound(fn, 0, len, true);
if (lowerBound == -1) { return []; }
const upperBound = this.binarySearchBound(fn, 0, len, false);
return items.slice(lowerBound, upperBound + 1);
}
binarySearchBound(compareFn: (a: LogItem) => number, left: number, right: number, isLowerBound: boolean): number {
const items = this.items;
if (items.length == 0) {
return -1;
}
if (isLowerBound && compareFn(items[0]) == 0) {
return 0;
} else if (!isLowerBound && compareFn(items[items.length - 1]) == 0) {
return items.length - 1;
}
let result = -1;
while (left <= right) {
const mid = (left + right) / 2;
const item = this.items[mid];
const compare = compareFn(item);
if (compare < 0) {
left = mid + 1;
} else if (compare > 0) {
right = mid - 1;
} else {
result = mid;
if (isLowerBound) {
right = mid - 1;
} else {
left = mid + 1;
}
}
}
return result;
}