Given an array of integers target. From a starting array, A consisting of all 1's, you may perform the following procedure :

• let x be the sum of all elements currently in your array.
• choose index i, such that 0 <= i < target.size and set the value of A at index i to x.
• You may repeat this procedure as many times as needed.

Return True if it is possible to construct the target array from A otherwise return False.

 

Example 1:

Input: target = [9,3,5]
Output: true
Explanation: Start with [1, 1, 1] 
[1, 1, 1], sum = 3 choose index 1
[1, 3, 1], sum = 5 choose index 2
[1, 3, 5], sum = 9 choose index 0
[9, 3, 5] Done

Example 2:

Input: target = [1,1,1,2]
Output: false
Explanation: Impossible to create target array from [1,1,1,1].

Example 3:

Input: target = [8,5]
Output: true

 public boolean isPossible(int[] A) {
        PriorityQueue<Integer> pq = new PriorityQueue<Integer>((a, b) -> (b - a));
        long total = 0;
        for (int a : A) {
            total += a;
            pq.add(a);
        }
        while (true) {
            int a = pq.poll();
            total -= a;
            if (a == 1 || total == 1)
                return true;
            if (a < total || total == 0 || a % total == 0)
                return false;
            a %= total;
            total += a;
            pq.add(a);
        }
    }

大佬牛逼啊

 

https://leetcode.com/problems/construct-target-array-with-multiple-sums/discuss/510256/JavaC%2B%2BPython-Backtrack-OJ-is-wrong 

整体逻辑是因为每次最大的数都是由sum得来的，所以逆推最大数 - 除此之外的sum，看最后数组是不是只剩下1来判断。

百分号的作用是避免重复计算

total==0是为[2]准备的

a % total == 0 为[4, 2]准备的