523. Continuous Subarray Sum

Link

Solution

這題基本上也是透一樣的概念去找到符合的subArray但是edge case有點多

   public boolean checkSubarraySum(int[] nums, int k) {
        int newK = (k > 0) ? k : -k;

        //continus 0 is all passed
        for(int i = 1; i < nums.length; i++){
                if(nums[i] == 0 && nums[i-1] == 0) return true;
        }
        
        //never possible
        if( newK == 0 ){
            return false;
        }
        
        //newK is 1 means any nums with length 2 is valid
        if( newK == 1 ){
            if(nums.length >= 2) return true;
            else return false;
        }
        
        HashMap<Integer, Integer> map = new HashMap<>();
        int sum = 0;
        map.put(0, -1);
        for(int i = 0; i < nums.length; i++){
            sum += nums[i];
            map.put(sum, i);
            int checkSum =  sum - newK;
            while( checkSum >= 0){
                if(map.get(checkSum) != null && i - map.get(checkSum) > 1) return true;
                checkSum -= newK;
            }
        }
        return false;
    }