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  • LeetCode

    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].

    解法一:利用sum之间的差暴力搜索

    class Solution {
        public int subarraySum(int[] nums, int k) {
            if (nums == null || nums.length == 0)
                return 0;
            int len = nums.length;
            int[] sum = new int[len+1];
            for (int i=0; i<len; i++)
                sum[i+1] = sum[i] + nums[i];
            int cnt = 0;
            for (int i=0; i<len; i++) {
                for (int j=i+1; j<=len; j++) {
                    if (sum[j] - sum[i] == k)
                        cnt ++;
                }
            }
            return cnt;
        }
    }

    解法二:利用map建立sum和出现次数cnt的映射关系

    class Solution {
        public int subarraySum(int[] nums, int k) {
            if (nums == null || nums.length == 0)
                return 0;
            Map<Integer, Integer> map = new HashMap<>();
            map.put(0, 1);
            int sum = 0, cnt = 0;
            for (int num : nums) {
                sum += num;
                cnt += map.getOrDefault(sum-k, 0);
                map.put(sum, map.getOrDefault(sum, 0)+1);
            }
            return cnt;
        }
    }

     第二种解法耗时更少

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  • 原文地址:https://www.cnblogs.com/wxisme/p/9791771.html
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