Prefix Sum + HashMap

1. hashmap<count，index>
2. map.containsKey()时更新最值不put，没有时put
3. 根据后面找前面，不用初始化prefix数组或hashmap

560. Subarray Sum Equals K

Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.

Example 1:
Input:nums = [1,1,1], k = 2
Output: 2

思路一致，不过对于hashmap的设计需要注意，跟其他题目不一样的地方是不是找最长的subarray，所以设计要不一样。

    public int subarraySum(int[] nums, int k) {
        // map to store prefix sum array
        // key is prefix sum value is index
        Map<Integer, Integer> map = new HashMap<>();
        int sum = 0;
        int res = 0;
        map.put(0, 1);
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            if (map.containsKey(sum - k)) {
                res += map.get(sum - k);
            }
            map.put(sum, map.getOrDefault(sum, 0) + 1);
        }
        return res;
    }

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.

余数放在map里
ex 【2，4】2%6 = 2 （4+2）% 6 = 0

    public boolean checkSubarraySum(int[] nums, int k) {
        if (nums == null || nums.length == 0) {
            return false;
        }
        Map<Integer, Integer> map = new HashMap<>();
        int sum = 0;
        // hold the position
        map.put(0, -1);
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            if (k != 0) {
                sum = sum % k;                
            }
            if (map.containsKey(sum)) {
                // make sure at least subarray has more than 2 elements
                if (i - map.get(sum) > 1) {
                    return true;
                }
            } else {
                map.put(sum, i);
            }
        }
        return false;
    }

325. Maximum Size Subarray Sum Equals k

Given nums = [1, -1, 5, -2, 3], k = 3,
return 4. (because the subarray [1, -1, 5, -2] sums to 3 and is the longest)

Subarray类题目需要想到前缀和数组，优化➕Hashmap
sum(i~j) = PrefixSum(j+1) - PrefixSum(i) PrefixSum[0] = 0

public int maxSubArrayLen(int[] nums, int k) {
          if(nums == null || nums.length == 0) return 0;
          int length = nums.length, sum = 0, maxSubLen = 0;
          //Using a hash map to store the sum of all the values before and include nums[i]
          Map<Integer, Integer> map = new HashMap();
          for(int i = 0; i < length; i++) {
              sum += nums[i];
              if(sum == k) {
                  maxSubLen = Math.max(maxSubLen, i + 1);
              } else if(map.containsKey(sum - k)) {
                  maxSubLen = Math.max(maxSubLen, i - map.get(sum - k));
              }
              if(!map.containsKey(sum)) {
                  map.put(sum, i);
              }
          }
          return maxSubLen;
      }

525. Contiguous Array

Given a binary array, find the maximum length of a contiguous subarray with equal number of 0 and 1.
给一个binary数组，找出最长的子数组，0，1数量相等

1 max初始化为0，如果（0，0，0，0）情况
2 contain时不put
3 用一个count变量记录就够了

    public int findMaxLength(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        int count = 0;
        int max = 0;
        Map<Integer, Integer> map = new HashMap<>();
        map.put(0, -1);
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 0) {
                count++;
            } else {
                count--;
            }
            if (map.containsKey(count)) {
                // if contains no put, get the leftmost 
                max = Math.max(max, i - map.get(count));
                
            } else {
                map.put(count, i);
            }
        }
        return max;
    }