Mar*_*ijn 8 database algorithm limit offset
这应该是一个有趣的挑战。我正在寻找一种尚不存在的算法(据我所知)
getFromDatabase(int page, int size)。getRecords(int offset, int limit)。不知何故,我们必须使用给定的offsetandlimit来检索只能由page和访问的匹配数据库记录size。显然,偏移/限制并不总是映射到单个页面/大小。挑战在于找到一种算法,使getFromDatabase检索所有记录的“理想”调用次数。该算法应考虑以下几个因素:
getFromDatabase都有一定的开销成本;尽量减少通话。我提出了以下算法:http : //jsfiddle.net/mwvdlee/A7J9C/ (JS 代码,但该算法与语言无关)。本质上它是以下伪代码:
do {
do {
try to convert (offset,limit) to (page,size)
if too much waste
lower limit by some amount
else
call `getDatabaseRecords()`
filter out waste records
increase offset to first record not yet retrieved
lower limit to last records not yet retrieved
} until some records were retrieved
} until all records are retrieved from database
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该算法的关键在于确定too much waste和some amount。但是这个算法不是最优的,也不能保证是完整的(很可能是,我只是无法证明)。
有没有更好的(已知的?)算法,或者我可以做的改进吗?有没有人对如何解决这个问题有好的想法?
正如@usr 所指出的,在这个问题的大多数变体中(无论是查询数据库、API 还是其他一些实体),最好尽可能减少调用次数,因为返回一些额外的行几乎总是更便宜而不是发出单独的电话。以下 PageSizeConversion 算法将始终找到返回尽可能少记录的单个调用(这正是它执行搜索的方式)。在数据集的开头 ( headWaste) 或结尾 ( tailWaste)可能会返回一些额外的记录,需要将数据集放入单个页面中。该算法在此处用 Javascript 实现,但很容易移植到任何语言。
function PageSizeConversion(offset, limit) {
var window, leftShift;
for (window = limit; window <= offset + limit; window++) {
for (leftShift = 0; leftShift <= window - limit; leftShift++) {
if ((offset - leftShift) % window == 0) {
this.pageSize = window;
this.page = (offset - leftShift) / this.pageSize;
this.headWaste = leftShift;
this.tailWaste = ((this.page + 1) * this.pageSize) - (offset + limit);
return;
}
}
}
}
var testData = [
{"offset": 0,"limit": 10,"expectedPage": 0,"expectedSize": 10,"expectedHeadWaste": 0,"expectedTailWaste": 0},
{"offset": 2,"limit": 1,"expectedPage": 2,"expectedSize": 1,"expectedHeadWaste": 0,"expectedTailWaste": 0},
{"offset": 2,"limit": 2,"expectedPage": 1,"expectedSize": 2,"expectedHeadWaste": 0,"expectedTailWaste": 0},
{"offset": 5,"limit": 3,"expectedPage": 1,"expectedSize": 4,"expectedHeadWaste": 1,"expectedTailWaste": 0},
{"offset": 3,"limit": 5,"expectedPage": 0,"expectedSize": 8,"expectedHeadWaste": 3,"expectedTailWaste": 0},
{"offset": 7,"limit": 3,"expectedPage": 1,"expectedSize": 5,"expectedHeadWaste": 2,"expectedTailWaste": 0},
{"offset": 1030,"limit": 135,"expectedPage": 7,"expectedSize": 146,"expectedHeadWaste": 8,"expectedTailWaste": 3},
];
describe("PageSizeConversion Tests", function() {
testData.forEach(function(testItem) {
it("should return correct conversion for offset " + testItem.offset + " limit " + testItem.limit, function() {
conversion = new PageSizeConversion(testItem.offset, testItem.limit);
expect(conversion.page).toEqual(testItem.expectedPage);
expect(conversion.pageSize).toEqual(testItem.expectedSize);
expect(conversion.headWaste).toEqual(testItem.expectedHeadWaste);
expect(conversion.tailWaste).toEqual(testItem.expectedTailWaste);
});
});
});
// load jasmine htmlReporter
(function() {
var env = jasmine.getEnv();
env.addReporter(new jasmine.HtmlReporter());
env.execute();
}());Run Code Online (Sandbox Code Playgroud)
<script src="https://cdn.jsdelivr.net/jasmine/1.3.1/jasmine.js"></script>
<script src="https://cdn.jsdelivr.net/jasmine/1.3.1/jasmine-html.js"></script>
<link href="https://cdn.jsdelivr.net/jasmine/1.3.1/jasmine.css" rel="stylesheet" />
<title>Jasmine Spec Runner</title>Run Code Online (Sandbox Code Playgroud)
这也许并不确切是什么@Martijn正在寻找,因为它有时有大量的浪费的结果。但在大多数情况下,它似乎是解决一般问题的好方法。