We define a harmonious array is an array where the difference between its maximum value and its minimum value is exactly 1.

Now, given an integer array, you need to find the length of its longest harmonious subsequence among all its possible subsequences.

Example 1:

Input: [1,3,2,2,5,2,3,7]
Output:
 
 5
Explanation: The longest harmonious subsequence is [3,2,2,2,3].

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題意

從一個數組中找到一個harmonious array，這個harmonious array定義為陣列的最大值和最小值之差等於1，現在需要找到最長的這種陣列。

思路

找到最長的就是要找到出現次數最多的，然後可以利用hashmap來統計某個數字的出現次數，然後再迴圈陣列，找每個數字n的次數，以及每個數字加1(n+1)後的數字出現次數，

然後比較結果，找到最大的即可。不需要找n-1的出現次數。

程式碼

public static int findLHS(int[] nums){
	Map<Long, Integer> map = new HashMap<>();
   	for (long num : nums) {
            map.put(num, map.getOrDefault(num, 0) + 1);
    	}
    	int result = 0;
    	for (long key : map.keySet()) {
       	    if (map.containsKey(key + 1)) {
            	result = Math.max(result, map.get(key + 1) + map.get(key));
            }
    	}
    	return result;
}
 

這裡有用到map.getOrDefault(args,default)方法，這個方法的作用是在map中找key等於args，如果找到返回value，如果沒找到則返回預設值default。

相似的方法putIfAbsent方法，這個方法也是在map中根據key找，如果找到，返回value，如果沒找到，插入這個key，返回null；

putIfAbsent

If the specified key is not already associated with a value (or is mapped to null) associates it with the given value and returns null, else returns the current value.

getOrDefault

Returns the value to which the specified key is mapped, or defaultValue if this map contains no mapping for the key.

If your goal is only to retrieve the value, then use getOrDefault. Else, if you want to set the value when it does not exist, use putIfAbsent.

我寫的另外一個方法

public static int findLHS(int[] nums) {
        int length = nums.length;
        if(nums == null || nums.length <=0)
        	return 0;
        int firstValue = nums[0];
        int i=1;
        int maxCount=0;
        while(i<length){
        	int count=0;
        	int j=0;
        	boolean flag = false;
        	while(j<length){
        		if( nums[j] == firstValue+1){        			        		
        			flag = true;
        			count++;
        		}
        		if(nums[j] == firstValue){
        			count++;
        		}
        		j++;
        	}
        	firstValue = nums[i];
        	if(count > maxCount && flag){
        		maxCount = count;
        	}        	
        	i++;        
        }
        	return maxCount;	                
    }

運行了幾個測試用例，返回值都是對的，但是leetCode提示超時