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
Note:
The length of the array is in range [1, 20,000].
The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].

fb: 只用返回true or false。第二题先用set做，后来让改用constant space， 就用了sliding window这样

subarray sum 问题常用hashmap, 存count 值和坐标, 动归的感觉啊

 fb:问了数组包含／不包含负数两种情况，

public int subarraySum(int[] nums, int k) {
        int sum = 0, result = 0;
        Map<Integer, Integer> preSum = new HashMap<>();
        preSum.put(0, 1);
        
        for (int i = 0; i < nums.length; i++) {
            sum += nums[i];
            if (preSum.containsKey(sum - k)) {
                result += preSum.get(sum - k);
            }
// 当加着加着出现两个一样的sum时, 要在他的value上加1, 因为可以有多个连续的串
            preSum.put(sum, preSum.getOrDefault(sum, 0) + 1);
        }
        
        return result;
}

要用 preSum.put(0, 1); 是得result 加的值可以来自map中的多个.

不能 if (sum == k) {

result++;
}

因为:

Input:[0,0,0,0,0,0,0,0,0,0] 0

Output:10

Expected:55

 

对比523. Continuous Subarray Sum, 0 的indice放-1, 0的count 放1;

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

　　对比subarray sum 

public class Solution {
    public boolean isIsomorphic(String s, String t) {
        if (s==null || t==null || s.length() != t.length()) return false;
        HashMap<Character, Character> map = new HashMap<Character, Character>();
        for (int i=0; i<s.length(); i++) {
            if (!map.containsKey(s.charAt(i))) {
                if (!(map.values().contains(t.charAt(i)))) {
                    map.put(s.charAt(i), t.charAt(i));
                }
                else return false;
            }
            else if (map.get(s.charAt(i)) != t.charAt(i)) return false;
        }
        return true;
    }
}