Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search circularly to find its next greater number. If it doesn't exist, output -1 for this number.
Example 1:
Input: [1,2,1] Output: [2,-1,2] Explanation: The first 1's next greater number is 2;
The number 2 can't find next greater number;
The second 1's next greater number needs to search circularly, which is also 2.
Note: The length of given array won't exceed 10000.
A:
就是搞一个 increasing stack. 最上面的最小。 唯一的tricky地方就是, 把nums 延展一倍。
class Solution { public: vector<int> nextGreaterElements(vector<int>& nums) { int n = nums.size(); vector<int> nums2(nums); nums2.insert(nums2.end(), nums.begin(), nums.end()); stack<int> minStack; // top value is always smallest vector<int> res(n); for(int i= nums2.size()-1;i>=0;i--){ int val = nums2[i]; while(not minStack.empty() and minStack.top() <= val){ minStack.pop(); } if(i<n){ res[i] = minStack.empty()?-1:minStack.top(); } minStack.push(val); } return res; } };
Errors:
一开始想nums2中,省掉最后一个, 然而当nums ==[] 时,会报错
nums2.insert(nums2.end(), nums.begin(), nums.end()-1);
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