leetcode 1354

问题

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:

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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:

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

Example 3:

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

Constraints:

  • N == target.length
  • 1 <= target.length <= 5 * 10^4
  • 1 <= target[i] <= 10^9

分析

分析可知问题输入的数组，数之间的顺序关系是无所谓的，[9,3,5] 和 [3,5,9] 的最终答案是一样的，并且每个目标数组的前一个状态的数组是确定的。一个数组的最后一步发生的变化一定是发生在了最大那个数字上。比如 [9,3,5] 最后一步发生的变化一定发生在了 9 上，因为要指定一个数变成整个数组的和，所以变化的位置变化之后一定存储的是数组中最大的那个数。于是我们可以从目标数递推判断是否可以变成全1的数组。如：[9,3,5] -> [1,3,5] -> [1,3,1] -> [1,1,1]。使用优先队列来维护数组中最大值的信息。

代码

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class Solution {
public:
    bool isPossible(vector<int>& target) {
        long long sum = 0;
        priority_queue<int, vector<int>, less<int>> q;
        for (int a: target) {
            q.push(a);
            sum += a;
        }
      	// q.top() == 1 则整个数组的元素都是1
        while (q.size() != 0 && q.top() != 1) {
            int x = q.top();
            q.pop();
            int y = x - (sum - x);  // 恢复原值
            if (y <= 0 || y >= x) return false;
            sum -= (x - y);
            q.push(y);
        }
        return true;
    }
};