[LeetCode] 560. Subarray Sum Equals K 子数组和为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:

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

 

这道题给了我们一个数组,让求和为k的连续子数组的个数,博主最开始看到这道题想着肯定要建立累加和数组啊,然后遍历累加和数组的每个数字,首先看其是否为k,是的话结果 res 自增1,然后再加个往前的循环,这样可以快速求出所有的子数组之和,看是否为k,参见代码如下:

 

解法一:

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        int res = 0, n = nums.size();
        vector<int> sums = nums;
        for (int i = 1; i < n; ++i) {
            sums[i] = sums[i - 1] + nums[i];
        }
        for (int i = 0; i < n; ++i) {
            if (sums[i] == k) ++res;
            for (int j = i - 1; j >= 0; --j) {
                if (sums[i] - sums[j] == k) ++res;
            }
        }
        return res;
    }
};

 

上面的求累加和的方法其实并没有提高程序的执行效率,跟下面这种暴力搜索的解法并没有什么不同,博主很惊奇 OJ 居然这么大度,让这种解法也能通过,参见代码如下:

 

解法二:

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        int res = 0, n = nums.size();
        for (int i = 0; i < n; ++i) {
            int sum = nums[i];
            if (sum == k) ++res;
            for (int j = i + 1; j < n; ++j) {
                sum += nums[j];
                if (sum == k) ++res;
            }
        }
        return res;
    }
};

 

论坛上大家比较推崇的其实是这种解法,用一个 HashMap 来建立连续子数组之和跟其出现次数之间的映射,初始化要加入 {0,1} 这对映射,这是为啥呢,因为解题思路是遍历数组中的数字,用 sum 来记录到当前位置的累加和,建立 HashMap 的目的是为了可以快速的查找 sum-k 是否存在,即是否有连续子数组的和为 sum-k,如果存在的话,那么和为k的子数组一定也存在,这样当 sum 刚好为k的时候,那么数组从起始到当前位置的这段子数组的和就是k,满足题意,如果 HashMap 中事先没有 m[0] 项的话,这个符合题意的结果就无法累加到结果 res 中,这就是初始化的用途。上面讲解的内容顺带着也把 for 循环中的内容解释了,这里就不多阐述了,有疑问的童鞋请在评论区留言哈,参见代码如下:

 

解法三:

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        int res = 0, sum = 0, n = nums.size();
        unordered_map<int, int> m{{0, 1}};
        for (int i = 0; i < n; ++i) {
            sum += nums[i];
            res += m[sum - k];
            ++m[sum]; 
        }
        return res;
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/560

 

类似题目:

Two Sum

Continuous Subarray Sum

Subarray Product Less Than K

Find Pivot Index

 

参考资料:

https://leetcode.com/problems/subarray-sum-equals-k/

https://leetcode.com/problems/subarray-sum-equals-k/discuss/102153/Basic-Java-solution

https://leetcode.com/problems/subarray-sum-equals-k/discuss/134689/Three-Approaches-With-Explanation

https://leetcode.com/problems/subarray-sum-equals-k/discuss/102106/Java-Solution-PreSum-%2B-HashMap

 

LeetCode All in One 题目讲解汇总(持续更新中...)


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posted @ 2017-05-04 22:46  Grandyang  阅读(21870)  评论(9编辑  收藏  举报
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