有點像Leetcode 68. Text Justification

問題
There are N items to be printed on receipt, the i-th item has Item[i] characters. A item must be printed on a single line, and can't change the item printing order.
Return the minimum number of character per line C, such that all the items can be printed exactly in L lines

input [1,2,3,4,5,6,7,8,9,10] L =5

Explain
L1 : 1,2,3,4,5
L2: 6,7
L3: 8
L4: 9
L5: 10

我現在寫出來邏輯因該是錯的，但不曉的怎麼改

public static int findMinimumNumber(int[] nums,int K){

        int minimum = 0;
        int totalNum = 0;
        int longest = 0;
        // calculate total # of characters
        for(int num : nums){
            longest = Math.max(longest,num);
            totalNum += num;
        }

        // find the longest word as Left, total # as Right
        // binary search, middle value divide totalNum

        int l = longest, r = totalNum;

        while(l < r){

            int mid = l + (r-l)/2;
            int line = totalNum / mid;

            // compare result to K
            // If result < K , update Left = mid+1; otherwise Right = mid;
            if(line == K){
                mid += K%mid;
                minimum = mid;
                break;
            }

            if(line < K){
                r = mid-1;
            }else{
                l = mid;
            }
        }

        return minimum;
    }

已经验证了K的话，更新的时候，l = mid + 1就好了，这个时候上面while的条件应该是l <= r。你这样写的话，二分查找是有可能死循环的。