💡 解题思路
- 📝 确定输入与输出
- 🔍 分析复杂度
- 🔨 复杂题目拆分 :严谨且完整 地拆分为更小的子问题(哈希表的使用场景)–(多总结)
- 💭 选择处理逻辑: 根据拆分后的子问题,总结并选择合适的问题处理思路
- 🔎 检查特殊情况:边界条件和特殊情况
- 🏁 返回结果
● 242.有效的字母异位词 (做统计)
class Solution {public boolean isAnagram(String s, String t) {if (s.length() != t.length()) {return false;}int[] table = new int[26];for (int i = 0; i < s.length(); i++) {table[s.charAt(i) - 'a']++;}for (int i = 0; i < t.length(); i++) {table[t.charAt(i) - 'a']--;if (table[t.charAt(i) - 'a'] < 0) return false;}return true;}
}
● 349. 两个数组的交集 (Set 去重)
class Solution {public static int[] intersection(int[] nums1, int[] nums2){if (nums1.length == 0 || nums2.length == 0) {return new int[0];}int len1 = nums1.length, len2 = nums2.length;Set<Integer> set = new HashSet<>(); // 确保不重复for (int num : nums1) {set.add(num);}List<Integer> res = new ArrayList<>(); // 动态数组记录结果for (int num : nums2) {if (set.contains(num)) {res.add(num);set.remove(num);}}int[] result = new int[res.size()];for (int i = 0; i < res.size(); i++) {result[i] = res.get(i);}return result;}
}
● 202. 快乐数(快慢指针的实际应用(变体))
class Solution {public boolean isHappy(int n) {if (n == 1 || getNext(n) == 1) return true;int slow = n;int fast = getNext(n);while (fast != 1 && slow != fast) {slow = getNext(slow);fast = getNext(getNext(fast));}return fast == 1;}private int getNext(int n) {int totalSum = 0;while (n > 0) {int d = n % 10;n = n / 10;totalSum += d * d;}return totalSum;}
}
● 1. 两数之和 (哈希表降低时间复杂度)
class Solution {public int[] twoSum(int[] nums, int target) {if (nums.length <= 1) {return new int[2];}HashMap<Integer, Integer> map = new HashMap<>();int len = nums.length;for (int i = 0; i < len; i++) {if (!map.containsKey(nums[i])) {map.put(target-nums[i], i);} else {return new int[] {i, map.get(nums[i])};}}return new int[2];}
}