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package leetcode_1To300;
import java.util.HashMap;
/**
* 本代码来自 Cspiration,由 @Cspiration 提供
* 题目来源:http://leetcode.com
* - Cspiration 致力于在 CS 领域内帮助中国人找到工作,让更多海外国人受益
* - 现有课程:Leetcode Java 版本视频讲解(1-900题)(上)(中)(下)三部
* - 算法基础知识(上)(下)两部;题型技巧讲解(上)(下)两部
* - 节省刷题时间,效率提高2-3倍,初学者轻松一天10题,入门者轻松一天20题
* - 讲师:Edward Shi
* - 官方网站:https://cspiration.com
* - 版权所有,转发请注明出处
*/
public class _325_MaximumSizeSubarraySumEqualsk {
/**
* 325. Maximum Size Subarray Sum Equals k
* 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)
[1, -1, 5, -2, 3] k = 3
1, 0, 5, 3, 6 k = 3
time : O(n)
space : O(n)
* @param nums
* @param k
* @return
*/
public int maxSubArrayLen(int[] nums, int k) {
if (nums == null || nums.length == 0) return 0;
int res = 0;
HashMap<Integer, Integer> map = new HashMap<>();
map.put(0, -1);
for (int i = 1; i < nums.length; i++) {
nums[i] += nums[i - 1];
}
for (int i = 0; i < nums.length; i++) {
if (map.containsKey(nums[i] - k)) {
res = Math.max(res, i - map.get(nums[i] - k));
}
if (!map.containsKey(nums[i])) {
map.put(nums[i], i);
}
}
return res;
}
}