Leetcode-523. Continuous Subarray Sum

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。这次比赛略无语，没想到前3题都可以用暴力解。

博客链接：mcf171的博客

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Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.

Example 1:

Input: [23, 2, 4, 6, 7],  k=6
Output: True
Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.

Example 2:

Input: [23, 2, 6, 4, 7],  k=6
Output: True
Explanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.

Note:

1. The length of the array won't exceed 10,000.
2. You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

public class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        
        Set sums = new HashSet();
        boolean flag = false;
        if(k == 0){
            for(int i = 0 ; i < nums.length - 1; i ++){
                if(nums[i] == 0 && nums[i] == nums[i + 1]) {flag = true; break;}
            }
        }else{
            for(int item : nums){
                    Set temp = new HashSet();
                    Iterator it = sums.iterator();
                    while(it.hasNext()){
                        int sum = it.next() + item;
                        if(sum % k == 0 || sum == 0) {flag = true; break;}
                        temp.add(sum);
                    }
                    temp.add(item);
                    sums = temp;
                if(flag) break;
            }
        }
        return flag;
    }
}

在论坛看到一个解法，惊呆了，因为是非负数，遍历数组一遍一直加，并且记录 取mod k的结果，只要出现过，判断一下位置即可返回结果。

public class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
    Map map = new HashMap(){{put(0,-1);}};;
    int runningSum = 0;
    for (int i=0;i 1) return true;
        }
        else map.put(runningSum, i);
    }
    return false;
}
}