LeetCode数组Subarray求和问题

关于我的 Leetcode 题目解答，代码前往 Github：https://github.com/chenxiangcyr/leetcode-answers

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LeetCode题目：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].

以下代码的时间负责度为O(n^2)：

class Solution {
    public int subarraySum(int[] nums, int k) {
        if(nums == null || nums.length == 0) return 0;
        
        // sum[i]记录前i个元素的和
        int[] sum = new int[nums.length];
        sum[0] = nums[0];
        for(int i = 1; i < nums.length; i++) {
            sum[i] = sum[i - 1] + nums[i];
        }
        
        int count = 0;
        // 遍历各个range
        for(int i = 0; i < nums.length; i++) {
            for(int j = i; j < nums.length; j++) {
                if((sum[j] - (i == 0 ? 0 : sum[i - 1])) == k) {
                    count++;
                }
            }
        }
        
        return count;
    }
}

以下代码的时间负责度为O(n)：

class Solution {
    public int subarraySum(int[] nums, int k) {
        int count = 0;
        int sum = 0;
        
        // key: 累计和sum, value: 从idx = 0 开始有多少种方式可以累积到该sum
        HashMap <Integer, Integer> map = new HashMap < > ();
        map.put(0, 1);
        
        for (int i = 0; i < nums.length; i++) {
            sum = sum + nums[i];
            
            if (map.containsKey(sum - k)) {
                count = count + map.get(sum - k);
            }
            
            map.put(sum, map.getOrDefault(sum, 0) + 1);
        }
        
        return count;
    }
}

LeetCode题目：325. Maximum Size Subarray Sum Equals k
Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn't one, return 0 instead.

Note:
The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range.

Example 1:

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)

Example 2:

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

Follow Up:

• Can you do it in O(n) time?

以下代码的时间负责度为O(n^2)：

class Solution {
    public int maxSubArrayLen(int[] nums, int k) {
        if(nums == null || nums.length == 0) return 0;
        
        // sum[i]记录前i个元素的和
        int[] sum = new int[nums.length];
        sum[0] = nums[0];
        for(int i = 1; i < nums.length; i++) {
            sum[i] = sum[i - 1] + nums[i];
        }
        
        int max = 0;
        // 遍历各个range
        for(int i = 0; i < nums.length; i++) {
            for(int j = i; j < nums.length; j++) {
                if((sum[j] - (i == 0 ? 0 : sum[i - 1])) == k) {
                    max = Math.max(max, j - i + 1);
                }
            }
        }
        
        return max;
    }
}

以下代码的时间负责度为O(n)：

class Solution {
    public int maxSubArrayLen(int[] nums, int k) {
        
        int max = 0;
        int sum = 0;
        
        // key: 累计和sum, value: 从idx = 0 开始第一次累积到该sum 的索引位置
        // 为什么保存第一次的位置，因为是求 max size
        HashMap <Integer, Integer> map = new HashMap < > ();
        map.put(0, -1);
        
        for (int i = 0; i < nums.length; i++) {
            sum = sum + nums[i];
            
            if (map.containsKey(sum - k)) {
                max = Math.max(max, i - map.get(sum - k));
            }
            
            if (!map.containsKey(sum)) {
                map.put(sum, i);
            }
        }
        
        return max;
    }
}

LeetCode题目：[53. Maximum Subarray]
(https://leetcode.com/problems/maximum-subarray/description/)
Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.

以下代码的时间负责度为O(n^2)：

class Solution {
    public int maxSubArray(int[] nums) {
        int max = Integer.MIN_VALUE;
        
        for(int i = 0; i < nums.length; i++) {
            int sum = 0;
            for(int j = i; j < nums.length; j++) {
                sum = sum + nums[j];
                
                max = Math.max(max, sum);
            }
        }
        
        return max;
    }
}

以下代码的时间负责度为O(n)：

class Solution {
    public int maxSubArray(int[] nums) {
        // dp[i] means the maximum subarray ending with A[i]
        int[] dp = new int[nums.length];
        
        // init
        dp[0] = nums[0];
        int max = dp[0];
        
        for(int i = 1; i < nums.length; i++){
            dp[i] = nums[i] + (dp[i - 1] > 0 ? dp[i - 1] : 0);
            
            max = Math.max(max, dp[i]);
        }
        
        return max;
    }
}