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function logArray(arr) { | |
arr.forEach((value, index) => console.log(`${index.toString(2).padStart(Math.log2(arr.length), '0')}: ${value}`)) | |
} | |
function newArray(len) { | |
if (typeof Uint8Array !== 'undefined') { | |
return new Uint8Array(len) | |
} | |
return new Array(len) | |
} | |
function checkEner(items, requirements) { | |
var n = requirements.length | |
var len = 1 << n | |
var arr = newArray(len) | |
// 第 i 条不等式。 | |
// 假设 n == 4,即有 4 种元需求: | |
// i 为从 0000 到 1111,从右到左分别称为第 1,2,3,4 位 | |
// 当 i == 0110 时,代表不等式左边为第 2、3 个集合的并集 | |
for (var i = 0; i < len; i++) { | |
var filters = [] | |
for (var j = 0; j < n; j++) { | |
if (i & (1 << j)) { | |
filters.push(requirements[j].filter) | |
// 不等式的右边(先把需求数加上去,满足的时候减 1,减至 0 则满足该不等式) | |
arr[i] += requirements[j].count | |
} | |
} | |
// 遍历能量区,满足条件的减 1 | |
for (var j = 0; j < items.length; j++) { | |
var item = items[j] | |
if (filters.some(function (filter) { | |
return filter(item) | |
})) { | |
// 减至 0,该不等式成立 | |
if (!--arr[i]) break | |
} | |
} | |
if (arr[i]) return false | |
} | |
logArray(arr) | |
return true | |
} | |
var items = [ | |
'红蓝', | |
'红', | |
'蓝', | |
'蓝', | |
'蓝', | |
'蓝', | |
] | |
var requirements = [ | |
{ | |
count: 1, | |
filter: item => item.includes('红'), | |
}, | |
{ | |
count: 2, | |
filter: item => item.includes('蓝'), | |
}, | |
] | |
checkEner(items, requirements) |
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