169. Majority Element
2017-05-08 18:45
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这道题的意思是,假设数组长度为n,让找出现次数大于[n/2]的元素是哪个,这样的元素只会有一个,所以只要找到出现次数大于[n/2]的元素一定是要找的答案。这里用了一个hashmap,value的更新方法是map.get(nums[i])+1,真的是要记住了,因为这一点代码调了好久==
还看了一个方法只用了2行java程序,排完序之后找到下标为nums.length/2的元素即可:
Arrays.sort(nums);
int len = nums.length; //int count=1; int k=0; HashMap<Integer,Integer> map =new HashMap<Integer,Integer>(); for(int i=0;i<len;i++){ if(!map.containsKey(nums[i])){ //count=1; map.put(nums[i],1); } else map.put(nums[i],map.get(nums[i])+1); //System.out.println(map.get(nums[i])); if(map.get(nums[i])>len/2){ k = nums[i]; return k; } } return k; }
还看了一个方法只用了2行java程序,排完序之后找到下标为nums.length/2的元素即可:
Arrays.sort(nums);
return nums[nums.length/2];
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