Merge pull request #2001 from BaeKwangho/main

[dev_qorh] WEEK1 solutions

Author: BaeKwangho

house-robber/Baekwangho.ts

@@ -0,0 +1,73 @@
+/** 1차 시도
+function rob(nums: number[]): number {
+  const jumpArray = []
+
+    if (nums.length<3) {
+        nums.sort((a,b)=> b-a)
+        return nums[0]
+    }
+
+  for (let i = 0; i<=nums.length-3; i++) {
+    jumpArray.push(nums[i]+nums[i+2])
+  }  
+
+  let maximum = 0
+
+  for (let i = 0; i<jumpArray.length-2; i+=3) {
+    if(jumpArray[i] >= jumpArray[i+1]) {
+        maximum += jumpArray[i]
+    }else{
+        maximum += jumpArray[i+1]
+    }
+  }
+
+  return maximum
+};
+*/
+
+/** 2차 시도, 풀이 참고
+f(nums) => nums 를 털어 구할 수 있는 가장 큰 금액
+첫집부터 털었을 때: nums[0] + f(nums[2:])
+둘째집부터 털었을 때: f(nums[1:])
+
+첫 시작
+f(nums) = MAX(nums[0] + f(nums[2:]), f(nums[1:]))
+
+function rob(nums: number[]): number {
+    function f (num: number) {
+        if(nums.length - 1 < num) {
+            return 0
+        }
+
+        const first = nums[num] + f(num + 2)
+        const second = f(num + 1)
+        return first >= second ? first : second
+    }
+
+    return f(0)
+}
+ */
+
+function rob(nums: number[]): number {
+  const resultMap = new Map();
+
+  function f(num: number) {
+    if (resultMap.has(num)) {
+      return resultMap.get(num);
+    }
+
+    if (nums.length - 1 < num) {
+      resultMap.set(num, 0);
+    } else {
+      const first = nums[num] + f(num + 2);
+      const second = f(num + 1);
+
+      const result = first >= second ? first : second;
+      resultMap.set(num, result);
+    }
+
+    return resultMap.get(num);
+  }
+
+  return f(0);
+}

two-sum/Baekwangho.ts

@@ -0,0 +1,14 @@
+/**
+ * 가장 단순한 O(n^2) 로 풀이하였습니다
+ */
+function twoSum(nums: number[], target: number): number[] {
+  for (let i = 0; i < nums.length; i++) {
+    for (let j = i + 1; j < nums.length; j++) {
+      if (nums[i] + nums[j] === target) {
+        return [i, j];
+      }
+    }
+  }
+
+  return [];
+}