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class Solution {
    public int maxLength(int[] ribbons, int k) {
        int left = 1, right = (int) 1e5, ans = 0;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (isCuttable(ribbons, mid, k)) {
                ans = mid;
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return ans;
    }

    private boolean isCuttable(int[] ribbons, int length, int k) {
        int count = 0;
        for (int ribbon : ribbons) {
            count += ribbon / length;
        }
        return count >= k;
    }
}

跟我想的一样. binary search. 我们需要一个method来判断当前的length是否能够取到, 取到的话记录一下然后往右走, 取不到就往左走.

时间复杂度: O(nlog(max(ribbons)))
空间复杂度: O(1)