# [front end algorithm] 1046. The weight of the last stone, violent solution

Time：2021-6-14

There is a pile of stones, and the weight of each stone is a positive integer.

Each round, choose two of the heaviest stones, and then smash them together. Suppose the weight of the stone is x and Y respectively, and x < = y. The possible results of comminution are as follows:

• If x = = y, both stones will be completely crushed;
• If x= y. Then the stone with weight x will be completely crushed, and the new weight of the stone with weight y is y-x.
• In the end, there’s only one stone left. Return the weight of this stone. If there is no stone left, return 0.
``````Example:
Input: [2,7,4,1,8,1]
Output: 1
Explanation:
First select 7 and 8 to get 1, so the array is converted to [2,4,1,1,1],
Then select 2 and 4 to get 2, so the array is converted to [2,1,1,1],
Then 2 and 1, get 1, so the array is converted to [1,1,1],
Finally, select 1 and 1, get 0, and finally convert the array to [1], which is the weight of the last remaining stone.``````

Tips:

• 1 <= stones.length <= 30
• 1 <= stones[i] <= 1000

How to solve the problem:

• 1. Sort
• 2. The largest minus the weight of the second largest stone
• 3. Delete the second item and change the subtraction value of the first item;
• 4. Reorder until the array length is less than 1

Problem solving code

``````/**
* @param {number[]} stones
* @return {number}
*/
var lastStoneWeight = function(stones) {
while (stones.length > 1) {
stones.sort((a, b) => b - a);
stones[0] = stones[0] - stones[1];
stones.splice(1, 1);
}
return stones.length === 0 ? 0 : stones[0]
};``````

results of enforcement

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