import java.util.HashMap;
import java.util.Map;public class _560_subarray_sum_equals_k {/* leetcode 560 号算法题：和为K的子数组给定一个整数数组和一个整数 k，你需要找到该数组中和为 k 的连续的子数组的个数。输入:nums = [1,1,1], k = 2输出: 2输入:nums = [0,1,-1,1,1,2], k = 0输出: 4- 数组的长度为 [1, 20,000]。- 数组中元素的范围是 [-1000, 1000] ，且整数 k 的范围是 [-1e7, 1e7]。*/public int subarraySum(int[] nums, int k) {Map<Integer, Integer> map = new HashMap<>();map.put(0, 1);int prefixSum = 0;int res = 0;for (int i = 0; i < nums.length; i++) {prefixSum += nums[i];int diff = prefixSum - k;// 哈希查找if (map.containsKey(diff)) {res += map.get(diff);}map.put(prefixSum, map.getOrDefault(prefixSum, 0) + 1);}return res;}// 前缀和 + 哈希查找public int subarraySum3(int[] nums, int k) {int[] prefixSum = new int[nums.length + 1];prefixSum[0] = 0;for (int i = 1; i <= nums.length; i++) {prefixSum[i] = prefixSum[i - 1] + nums[i -1];}Map<Integer, Integer> map = new HashMap<>();int res = 0;for (int i = 0; i < prefixSum.length; i++) {int diff = prefixSum[i] - k;// 哈希查找if (map.containsKey(diff)) {res += map.get(diff);}map.put(prefixSum[i], map.getOrDefault(prefixSum[i], 0) + 1);}return res;}// 前缀和 + 线性查找public int subarraySum2(int[] nums, int k) {int[] prefixSum = new int[nums.length + 1];prefixSum[0] = 0;for (int i = 1; i <= nums.length; i++) {prefixSum[i] = prefixSum[i - 1] + nums[i -1];}int res = 0;for (int i = 0; i <= nums.length; i++) {int diff = prefixSum[i] - k;for (int j = 0; j < i; j++) {if (prefixSum[j] == diff) res++;}}return res;}// 暴力解法public int subarraySum1(int[] nums, int k) {int res = 0;for (int i = 0; i < nums.length; i++) {for (int j = i; j < nums.length; j++) {int sum = 0;for (int m = i; m <= j; m++) {sum += nums[m];}if (sum == k) res++;}}return res;}// 第二种思路 暴力优化public int subarraySum5(int[] nums, int k) {int res = 0;for (int i = 0; i < nums.length; i++) {int sum = 0;for (int j = i; j >= 0; j--) {sum += nums[j];if (sum == k) res++;}}return res;}}