Leetcode 169 Majority Element (main element) (vector, map)

translation

给定一个长度为n的数组，找出主要的元素。所谓主要的元素是指的出现次数超过? n/2 ?次的元素。你可以假定这个数组是非空的，并且“主要元素”一定是存在的。

Original

anoftheelementelementtheelement that appears more than ? n/2the array is non-emptyandtheelementinthe array.

Analysis

Since test instructions said that the main element is certain, I drilled the loophole, did not intend to ask what n/2 this thing, directly is the longest in all sequences.

1,2,2,3,3,3,3,3,4,4-->1-12-23-54-2

See the key value pair, think of the basic map, there are some complicated containers I use is not very familiar with ...

In this two of the code below, I first defined element elements and then used range-for to facilitate all elements from the nums:

1）如果在map中找不到这个元素，就添加进去；2）如果找到了，将其出现的个数加上1。如果当前的长度大于最大的长度，则进行一些列操作，并将max设置为当前的n，以便于后面的返回。

After debugging, it is found that the nums length is 1, so add a judgment at the beginning.

intMajorityelement ( vector<int>& Nums) {if(nums.size () = =1)returnnums[0]; map<int, int>ElementintMax =0, MaxLen =0; for(AutoN:nums) {if(Element.find (n) = = Element.end ()) {Element.insert ( map<int, int>:: Value_type (N,1)); }Else{Element[n] + =1;if(Element[n] >= maxlen)                {maxlen = Element[n];            max = n; }        }    }returnMax;}

Code

classSolution { Public:intMajorityelement ( vector<int>& Nums) {if(nums.size () = =1)returnnums[0]; map<int, int>ElementintMax =0, MaxLen =0; for(AutoN:nums) {if(Element.find (n) = = Element.end ()) {Element.insert ( map<int, int>:: Value_type (N,1)); }Else{Element[n] + =1;if(Element[n] >= maxlen)                    {maxlen = Element[n];                max = n; }            }        }returnMax }};

Leetcode 169 Majority Element (main element) (vector, map)