2019/10/15

Author: rchen102

总结/Array/Array.MD

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+## 1 Two Sum
+
+### Solution1: Brute force
+T: O(n^2) S: O(1)
+
+```java
+class Solution {
+    public int[] twoSum(int[] nums, int target) {
+        int[] res = new int[2];
+        for (int i = 0; i < nums.length; i++) {
+            for (int j = i + 1; j < nums.length; j++) {
+                if (nums[i] + nums[j] == target) {
+                    res[0] = i;
+                    res[1] = j;
+                    return res;
+                }
+            }
+        }
+        return res;
+    }
+}
+```
+
+### Solution2: HashMap
+T: O(n) S: O(n)
+
+```java
+class Solution {
+    public int[] twoSum(int[] nums, int target) {
+        int[] res = new int[2];
+        Map<Integer, Integer> map = new HashMap<>();
+        for (int i = 0; i < nums.length; i++) {
+            if (map.containsKey(target - nums[i])) {
+                res[0] = map.get(target - nums[i]);
+                res[1] = i;
+                return res;
+            }
+            map.put(nums[i], i);
+        }
+        return res;
+    }
+}
+```
+
+```py
+class Solution:
+    def twoSum(self, nums, target):
+        dic = {}
+        for i in range(len(nums)):
+            if (target - nums[i]) in dic:
+                return [dic[target - nums[i]], i]
+            else:
+                dic[nums[i]] = i
+```

总结/Array/Leetcode 1.MD

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+## 15 3Sum
+
+### Solution1: