K 次串联后最大子数组之和 JS

 

  Kadnae 解法:

var kConcatenationMaxSum = function (arr, k) {
    let sum = arr[0];
    let newK = k > 2 ? 2 : k;
    let pre = 0;
    let an = 0;
    for (let i = 0; i < newK; i++) {
        if (pre > 0) {
            pre = pre + arr[i % arr.length];
        } else {
            pre = arr[i % arr.length];
        }
        an = an > pre ? an : pre;
        if (i < arr.length) {
            sum += arr[i];
        }
    }
    while (sum > 0 && --k >= 2) {
        an = an + sum;
    }
    return an;
};

原文地址:https://www.cnblogs.com/niuyourou/p/13430792.html