[codility]equi

最新推荐文章于 2022-03-07 13:31:19 发布

原创最新推荐文章于 2022-03-07 13:31:19 发布 · 1.1k 阅读

0 ·

CC 4.0 BY-SA版权

codility 专栏收录该内容

3 篇文章

订阅专栏

本文介绍了一种算法，用于在一个整数数组中找到一个索引P，使得该索引左侧所有元素之和等于右侧所有元素之和。通过遍历数组并使用前缀和的概念，该算法能在O(N)的时间复杂度内解决问题。

Find an index in an array such that its prefix sum equals its suffix sum.

A[0] + A[1] + ... + A[P−1] = A[P+1] + ... + A[N−2] + A[N−1].

A[0] = -7   A[1] =  1   A[2] = 5
A[3] =  2   A[4] = -4   A[5] = 3
A[6] =  0

P = 3 is an equilibrium index of this array, because:

A[0] + A[1] + A[2] = A[4] + A[5] + A[6]

P = 6 is also an equilibrium index, because:

A[0] + A[1] + A[2] + A[3] + A[4] + A[5] = 0

the function may return 3 or 6, as explained above.

Complexity:

expected worst-case time complexity is O(N);
expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

function solution($A) {
    // write your code in PHP5
    
    $sum=0;
    for($i=0;$i<count($A);$i++){    
        $sum+=$A[$i];
    }
    
    $left=0;
    
    for($i=0;$i<count($A);$i++){
        if($left==$sum-$A[$i]-$left){
            return $i;
        }
        $left+=$A[$i];    
    }
    
    return -1;
    
}