[LeetCode] [Easy] [JS] 1046. Last Stone Weight

 

1. 문제 및 예시

 

We have a collection of stones, each stone has a positive integer weight.

Each turn, we choose the two heaviest stones and smash them together.  Suppose the stones have weights x and y with x <= y.  The result of this smash is:

• If x == y, both stones are totally destroyed;
• If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x.

At the end, there is at most 1 stone left.  Return the weight of this stone (or 0 if there are no stones left.)

 

 

Example 1:
Input: [2,7,4,1,8,1]
Output: 1
Explanation: 
We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of last stone.

 

 

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

 

 

 

2. 풀이

 

/**
 * @param {number[]} stones
 * @return {number}
 */
var lastStoneWeight = function(stones) {
  while(stones.length > 1){
    let s1 = Math.max(...stones);
    stones.splice(stones.indexOf(s1), 1);
    let s2 = Math.max(...stones);
    stones.splice(stones.indexOf(s2), 1);
    if(s1 != s2) {
        stones.push(s1-s2);
    }
  }
  return (stones.length > 0) ? stones[0] : 0;
};

 

 

 

3. 결과

70 / 70 test cases passed.
Status: Accepted
Runtime: 96 ms
Memory Usage: 33.8 MB