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:

  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].

题目含义:求和为k的连续子数组的个数

 思路:最直观的想法是遍历数组并依次加当前位置的数字,同时用数组preSum记录下当前位置之前所有数字的相加和,这样下标[i, j)之间的数字之和就可以用preSum[j]-preSum]来计算,然后通过双层循环,遍历所有情况来统计满足条件的子数组个数

 1     public int subarraySum(int[] nums, int k) {
 2         int[] preSum = new int[nums.length];  //dp数组,位置i上存储从0到i个元素总和
 3         preSum[0] = nums[0];  
 4         for (int i = 1; i < nums.length; i++) preSum[i] = preSum[i - 1] + nums[i];
 5         int result = 0;
 6         for (int i = 0; i < preSum.length; i++) {
 7             if (preSum[i] == k) result++;
 8             for (int j = i + 1; j < preSum.length; j++) {
 9                 if (preSum[j] - preSum[i] == k) result++;   //线段2-线段1==长度k
10             }
11         }
12         return result;
13     }

 

posted @ 2017-10-16 19:26  daniel456  阅读(152)  评论(0编辑  收藏  举报