Question

有一个数组：[1,1,1,1,2,3,4]，计算数组中所有相同差值的集合 ※数组中每个集合的差值不能相同

Way to count one set have same difference:
-1,0,1  -> count = 1 with diff = 1
1,1,1   -> count = 1 with diff = 0
1,2,3,4 -> count = 3 with diff = 1 ([1,2,3],[1,2,3,4],[2,3,4])
0,1     -> not count
0,1,3   -> not count
0,1,1,2 -> not count , althought it has 0,1 and 1,2 with diff = 1 but it is not continous

As the above array  [1,1,1,1,2,3,4] count = 6
There are 6 sets has the same difference value:
 [1,1,1],[1,1,1,1],[1,1,1] with difference = 0
 [1,2,3],[1,2,3,4],[2,3,4] with difference = 1

这是我的暴力手段O（n2）我可以结合使用DP来更有效地改善它吗？

 function run(pArray){
      var count = 0;
      var length = pArray.length;
      for(var i = 0;i<length - 2;i++){          
        var nextIdx = i + 1;
        var diff = pArray[nextIdx]-pArray[i];          
        while(true){
          nextIdx ++;
        
          if(nextIdx > length - 1){
            break;
          }    
          if(pArray[nextIdx] - pArray[nextIdx - 1] !== diff){
            break;
          }else{
            count ++;
          }
        }
    }
    
    console.log(count);
    }
    
    run([1,1,1,1,2,3,4]);

Answer 1

根据您的最新评论，似乎您确实关心数组中元素的顺序。因此，不需要对其进行排序，在那种情况下，集合的数量可以像[1,2,1,2,1]那样评估为零。

在这种情况下，可以在O(n)中完成。

想法是：

如果找到匹配的差异，则递增一个计数变量，更新结果作为结果=结果+计数。

举个例子[1,1,1,1]。您无需考虑索引0,1。

采用索引2来查看arr [2]-arr [1] == arr [1]-arr [0]，然后递增从0到1计数，然后将其添加到结果中。

采用索引3来查看arr [3]-arr [2] == arr [2]-arr [1]，是的等于，然后将count增加到2，将其相加。现在结果= 1 + 2 = 3。

只要您知道满足以下条件的元素，您就会知道条件arr [i]-arr [i-1] == arr [i-1]-arr [i-2]您不断添加下一个更高的值。因此结果= 1 + 2 + 3 + .... i-2依此类推给定索引i

如果找不到匹配的差异，请将计数变量重置为零然后重新开始。

注意：我不太熟悉Javascript，但是看起来像这样：

<div>
    <script>
      function countSets(arr) {
        if (arr === undefined || arr.length < 3) return 0;
        let count = 0;
        let res = 0;
        for (let i = 2; i < arr.length; ++i) {
          if (arr[i] - arr[i - 1] === arr[i - 1] - arr[i - 2]) {
            count++;
          } else {
            count = 0;
          }
          res += count;
        }
        return res;
      }
      console.log("[1, 1, 1] -> " + countSets([1, 1, 1]))
      console.log("[1, 1, 1, 2, 2, 2] -> " + countSets([1, 1, 1, 2, 2, 2]))
      console.log("[1, 1, 1, 2, 3, 4] -> " + countSets([1, 1, 1, 2, 3, 4]))
      console.log("[1, 1, 2, 3, 4, 4, 5, 5] -> " + countSets([1, 1, 2, 3, 4, 4, 5, 5]))
      console.log("[1, 1, 1, 2, 3, 4, 5, 6] -> " + countSets([1, 1, 1, 2, 3, 4, 5, 6]))
      console.log("[1, 2, 1, 2, 1] -> " + countSets([1, 2, 1, 2, 1]))
    </script>
</div>

计数集在整数数组中具有相同的差异

1 个答案: