add to list
description
submission
solutions
⌊ n/2 ⌋times.
you may assume that the array is non-empty and the majority element always exist in the array.
演算法分析：結合本週所講的divide and conquer,先嘗試將該問題分解為兩個小問題：將該陣列分為前後兩個大小相等或者相差乙個元素的陣列，分別尋找這兩個陣列中出現次數最多的元素（同理對這兩個陣列繼續進行分解，直到只有乙個元素為止），對於這兩個分別最大的元素，1如果它們相等，則就是要找的元素；2 它們不相等，則遍歷對應的陣列比較它們誰出現的次數更多（複雜度為o(n) )。遞推式為t(n) =2t(n/2) +2o(n),由主定理可知複雜度為o(nlogn)。**如下
class solution
private:
int majority(vector& nums,int begin,int end)
//if(num1>num2)
//return front;
//else
//return back;
return (num1>num2)?front:back;}}
}};
演算法設計與應用基礎作業第二週
53.maximum subarray descripyion find the contiguous subarray within an array containing at least one number which has the largest sum.for example,give...
演算法分析與設計課程作業第二週 1
上一周學了分而治之的演算法策略，這週就先找了一題貼了此標籤的題試試手。find the contiguous subarray within an array containing at least one number which has the largest sum.for example,g...
演算法設計與應用基礎 第七周
intersection of two arrays ii given two arrays,write a function to compute their intersection.example given nums1 1,2,2,1 nums2 2,2 return 2,2 note th...