LeetCode #1046 LastStoneWeight. Algorithm,알고리즘,LeetCode,Codefights,CodeSignal,코드파이트,코드시그널,예제,문제해결능력,example,c++,java,재귀,recursive,datastructure,techinterview,coding,코딩인터뷰,기술면접,연결리스트
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Last Stone Weight.
Memory Usage: 8.3 MB, less than 100.00% of C++ online submissions for Last Stone Weight.
LeetCode #1046
Q.
We have a collection of rocks, each rock has a positive integer weight.
Each turn, we choose the two heaviest rocks and smash them together. Suppose the stones have weights x and ywith 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.)
바위를 모아둔 것이 있는데, 각 바위들은 양의정수로 무게를 가지고 있다.
각 턴마다, 우리는 두개의 가장 무거운 바위들을 부딪힌다. 무게가 x, y 인 돌들을 가지고 있다고 가정하고, x <= y 라고 생각해보자. 부딪힌 결과는 다음과 같다 :
- 만약 x == y 라면, 돌 둘다 완전 깨진다.
- 만약 x != y 라면, x 는 완전히 깨지고, y 는 y-x 값만큼 남아있는다.
마지막에는, 1개의 돌이 남는다. 이 돌의 무게를 반환해라. (아니면 돌이 없으면 0 을 반환해라.)
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.
Note:
- 1 <= stones.length <= 30
- 1 <= stones[i] <= 1000
Process
// Process
//1. Input stones
//2. Iterate till stones length becomes 0
// 2.1. Sort stones (decending)
// 2.2. Check if [0] and [1] are the same or not
// 2.2.1. If so -> remove both
// 2.2.2. If not -> subtract bigger one to smaller one, and remove smaller one
//3. Return resultStone weight
// 처리과정
//1. 바위 배열을 입력받는다.
//2. 바위가 1개보다 적을 때까지 반복한다.
// 2.1. 내림차순 정렬한다.
// 2.2. [0] 번째와, [1] 번째 값이 같은지 다른지 비교해서
// 2.2.1. 같으면 -> 둘다 지운다.
// 2.2.2. 다르면 -> 큰값은 작은값 빼고, 작은값은 지운다.
//3. 돌 값 반환한다.
Code.. lemme see example code!!!
코드.. 예제코드를 보자!!!
class Solution {
public:
int lastStoneWeight(vector<int>& stones) {
if (stones.size() > 0) {
while (stones.size() > 1) {
sort(stones.begin(), stones.end(), greater<int>());
if (stones[0] == stones[1]) {
stones.erase(stones.begin());
stones.erase(stones.begin());
}
else {
stones[0] = stones[0] - stones[1];
stones.erase(stones.begin() + 1);
}
}
}
if (stones.size() > 0)
return stones[0];
else
return 0;
}
};
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