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TypeScript Immutable.OrderedSet類代碼示例

本文整理匯總了TypeScript中Immutable.OrderedSet的典型用法代碼示例。如果您正苦於以下問題:TypeScript OrderedSet類的具體用法?TypeScript OrderedSet怎麽用?TypeScript OrderedSet使用的例子?那麽, 這裏精選的類代碼示例或許可以為您提供幫助。


在下文中一共展示了OrderedSet類的5個代碼示例,這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚,您的評價將有助於係統推薦出更棒的TypeScript代碼示例。

示例1: it

 it('unions a set and an iterable and returns a set', () => {
   var s1 = Set([1,2,3]);
   var emptySet = Set();
   var l = List([1,2,3]);
   var s2 = s1.union(l);
   var s3 = emptySet.union(l);
   var o = OrderedSet([1,2,3]);
   var s4 = s1.union(o);
   var s5 = emptySet.union(o);
   expect(Set.isSet(s2)).toBe(true);
   expect(Set.isSet(s3)).toBe(true);
   expect(Set.isSet(s4) && !OrderedSet.isOrderedSet(s4)).toBe(true);
   expect(Set.isSet(s5) && !OrderedSet.isOrderedSet(s5)).toBe(true);
 });
開發者ID:Harishs84,項目名稱:immutable-js,代碼行數:14,代碼來源:Set.ts

示例2: List

      return prices.flatMap((price: number) => {
        let newBudget = budget - price; // If we buy this item, what is our new budget?

        // Remove items that are more than our budget and more than the item
        // under consideration.
        let newMenuItems = prices.filter(c => {
          let priceCeiling = Math.min(newBudget, price);
          return c <= priceCeiling;
        }) as OrderedSet<number>; // Should remain an OrderedSet after filter

        // No recursion if the item under consideration exactly zeroes our
        // budget.
        if (newBudget === 0) {
          let results = List([List([price])]);
          this.memo[hashed] = results;
          return results;
        };

        // Recursive call
        let recursive = this.computeHelper(newMenuItems, newBudget);

        // If recursion returned results, concat the item under consideration
        // onto each result and return that. If recursion didn't return results
        // return empty set.
        let results = recursive
          ? recursive.map((e: List<number>) => e.concat(price)).toSet()
          : OrderedSet([]);

        this.memo[hashed] = results;
        return results;
      });
開發者ID:Ethan826,項目名稱:tablexi-coding-challenge,代碼行數:31,代碼來源:knapsack.ts

示例3: it

 it('provides initial values in a mixed order', () => {
   var s = OrderedSet.of('C', 'B', 'A');
   expect(s.has('A')).toBe(true);
   expect(s.has('B')).toBe(true);
   expect(s.has('C')).toBe(true);
   expect(s.size).toBe(3);
   expect(s.toArray()).toEqual(['C','B','A']);
 });
開發者ID:Harishs84,項目名稱:immutable-js,代碼行數:8,代碼來源:OrderedSet.ts

示例4: toggleSetItem

function toggleSetItem(state: OrderedSet<string>, id: string): OrderedSet<string> {
  return state.includes(id) ? state.remove(id) : state.add(id);
}
開發者ID:threehams,項目名稱:reverie-client,代碼行數:3,代碼來源:uiReducer.ts

示例5: computeHelper

  private computeHelper(
    prices: OrderedSet<number>,
    budget: number): any { // actually Set<List<number>>
    let hashed = this.hashArgs(prices, budget);

    let memoizedResult = this.memo[hashed];
    if (typeof memoizedResult !== "undefined") {
      return memoizedResult;
    }


    // Base cases

    // If there are no prices, there can be no solution. Return empty set.
    if (prices.size === 0) {
      let results = Set([]);
      this.memo[hashed] = results;
      return results;

      // With one price, return empty set if price is not a factor of budget
      // or return a list of length budget / price filled with price.
      // E.g., for price 2 and budget 8, return List([2, 2, 2, 2]).
    } else if (prices.size === 1) {
      let onlyElement = prices.toList().get(0);
      if (budget % onlyElement === 0) {
        let results = Set([List(Array(budget / onlyElement).fill(onlyElement))]);
        this.memo[hashed] = results;
        return results;
      } else {
        let results = Set([]);
        this.memo[hashed] = results;
        return results;
      }

      // Recursive case. Divide-and-conquer algorithm compiles and filters
      // results by recurring on each price price, subtracting that price
      // from the budget and filtering the list of price prices to be less
      // than or equal to both the new budget and the current price. See
      // README for additional information.
    } else {
      return prices.flatMap((price: number) => {
        let newBudget = budget - price; // If we buy this item, what is our new budget?

        // Remove items that are more than our budget and more than the item
        // under consideration.
        let newMenuItems = prices.filter(c => {
          let priceCeiling = Math.min(newBudget, price);
          return c <= priceCeiling;
        }) as OrderedSet<number>; // Should remain an OrderedSet after filter

        // No recursion if the item under consideration exactly zeroes our
        // budget.
        if (newBudget === 0) {
          let results = List([List([price])]);
          this.memo[hashed] = results;
          return results;
        };

        // Recursive call
        let recursive = this.computeHelper(newMenuItems, newBudget);

        // If recursion returned results, concat the item under consideration
        // onto each result and return that. If recursion didn't return results
        // return empty set.
        let results = recursive
          ? recursive.map((e: List<number>) => e.concat(price)).toSet()
          : OrderedSet([]);

        this.memo[hashed] = results;
        return results;
      });
    }
  };
開發者ID:Ethan826,項目名稱:tablexi-coding-challenge,代碼行數:73,代碼來源:knapsack.ts


注:本文中的Immutable.OrderedSet類示例由純淨天空整理自Github/MSDocs等開源代碼及文檔管理平台,相關代碼片段篩選自各路編程大神貢獻的開源項目,源碼版權歸原作者所有,傳播和使用請參考對應項目的License;未經允許,請勿轉載。