[LeetCode]594.Longest Harmonious Subsequence
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题目大意:点击打开链接
思路:将数组每个元素和它出现的次数放入HashMap<Integer,Integer>
然后在keySet()里判断,每个key和比它大一的key+1出现次数之和,但前提得有key+1
注意:题目要求是最大最小之差exactly 1,因此如果给定数组[1,1,1,1]或者[1,3,5,7]就没有符合条件的result,result=0
public class Solution { public int findLHS(int[] nums) { if(nums==null || nums.length==0){ return 0; } Map<Integer,Integer> map=new HashMap<>(); for(int num:nums){ map.put(num,map.getOrDefault(num,0)+1); } int result=0; //result初始值为0的重要性 for(int key:map.keySet()){ if(map.containsKey(key+1)){ result=Math.max(result,map.get(key)+map.get(key+1)); } } return result; }}
public class Solution { public int findLHS(int[] nums) { //错误做法 if(nums==null || nums.length==0){ return 0; } Map<Integer,Integer> map=new HashMap<>(); for(int num:nums){ map.put(num,map.getOrDefault(num,0)+1); //原因:[2,3,2,2,2] if(map.containsKey(num+1)){ //当遍历到第二个,第三个,第四个2时都会在次数上累加1 map.put(num,map.get(num)+1); } } int result=0; for(Map.Entry<Integer,Integer> e:map.entrySet()){ result=Math.max(result,e.getValue()); } return result; }}
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