523. Continuous Subarray Sum

Problem Description：

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.

 题解

这题有较多的corner case，特别是k=0；

1、暴力O(n^2)

对每组序列A[i],A[i+1],...,A[j-1],A[j]求和，计算其和是否能被k整除

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        int n = nums.length;
        if(k == 0) {
            for(int i = 0; i < n - 1; i++) {
                if(nums[i] == 0 && nums[i + 1] == 0) return true;
            }
            return false;
        }
        for(int i = 0; i < n; i++) {
            int sum = nums[i];
            for(int j = i + 1; j < n; j++) {
                sum += nums[j];
                if(sum % k == 0) return true;
            }
        }
        
        return false;
    }
}

2、HashMap  时间复杂度 O(n)

sum[i]代表从开头当i位置的求和。该方法的思路是，若sum[i] % k == sum[j] % k 且 i > j + 1, 则存在一个子序列。

用HashMap<Integer, Integer> map， 其key为从开头位置开始到当前位置的和的mod k的值，value为当前的index。

采用此种算法要做map.put(0, -1)，来应对出现sum[i] % k == 0 的情况。

class Solution {
    public boolean checkSubarraySum(int[] nums, int k) {
        int n = nums.length;
        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, -1);
        int runningSum = 0;
        for(int i = 0; i < n; i++) {
            runningSum += nums[i];
            if(k != 0) runningSum %= k;
            if(map.containsKey(runningSum)) {
                if(i - map.get(runningSum) > 1) return true;
            } else map.put(runningSum, i);
        }
        return false;
    }
}