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].
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 }