【编程】【leetcode】523. Continuous Subarray Sum
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public class Solution { public boolean checkSubarraySum(int[] nums, int k) { int l=nums.length; //如果区分第一个元素不为k 那么就可以根据sum%k来判断是否返回true //反之 根据sum%k存在一定的缺陷 6 1 1 1 //因此为了统一 不在根据sum 而都根据余数是否有相等 int modsum=0; HashMap<Integer,Integer> hashMap=new HashMap<Integer, Integer>(); hashMap.put(0,-1);//之所以要放这个是为了防止出现1 1 k==2的这种情况 for (int i = 0; i <l ; i++) { modsum+=nums[i]; if(k!=0){ //可能是负数倍数 如果k==0 那么因为不会出现相同的余数 最后返回结果为false modsum%=k; } if (hashMap.get(modsum)!=null){ if (i-hashMap.get(modsum)>1){//不能等于1 因为可能出现23 2 6等类似情况 return true; }else{ continue; } }else{ hashMap.put(modsum,i); } } return false; }}
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