[LeetCode] 976. Largest Perimeter Triangle 最大周长的三角形
Given an array A
of positive lengths, return the largest perimeter of a triangle with non-zero area, formed from 3 of these lengths.
If it is impossible to form any triangle of non-zero area, return 0
.
Example 1:
Input: [2,1,2]
Output: 5
Example 2:
Input: [1,2,1]
Output: 0
Example 3:
Input: [3,2,3,4]
Output: 10
Example 4:
Input: [3,6,2,3]
Output: 8
Note:
3 <= A.length <= 10000
1 <= A[i] <= 10^6
这道题给了个正整数数组,让从中选三个数当作三角形的三条边,问能组成的三角形的最大周长是多少。因为要组成三角形,所以必须要满足两边之和大于第三边这一条性质,我们并不用去检测所有的组合情况,而是只要判断较短的两边之和是否大于最长的那条边就可以了。虽然这道是 Easy 题目,但是 OJ 仍然不让用暴力搜索法,遍历任意三条边是会超时的。所以只能想优化的解法,既然要周长最长,则肯定是选较大的数字先测比较好。这里就先给数组排个序,然后从末尾开始,每次取出三个数字,先检测能否组成三角形,可以的话直接返回周长,不行的话就继续往前取,若都不行的话,就返回0,参见代码如下:
class Solution {
public:
int largestPerimeter(vector<int>& A) {
sort(A.begin(), A.end());
for (int i = (int)A.size() - 1; i >= 2; --i) {
if (A[i] < A[i - 1] + A[i - 2]) {
return A[i] + A[i - 1] + A[i - 2];
}
}
return 0;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/976
类似题目:
参考资料:
https://leetcode.com/problems/largest-perimeter-triangle/
https://leetcode.com/problems/largest-perimeter-triangle/discuss/217972/C%2B%2B-4-lines-O(n-log-n)