add 3573 & 1859

Author: clingoram

javascript/LeetCode/Array/3573.js

@@ -0,0 +1,37 @@
+/**
+ * 3573. Best Time to Buy and Sell Stock V
+ * 
+ * 正常交易：i天買，j天賣(i < j)，利潤：proces[j] - prices[i]
+ * 炒短線：i天賣，j天買回來(i < j)，利潤：prices[i] - prices[j]
+ * 不能同一天買賣單張股票
+ * 回傳在交易k次後最多能賺多少
+ * 
+ * @param {number[]} prices 第i天的股價
+ * @param {number} k 最多只能交易k次
+ * @return {number}
+ */
+var maximumProfit = function(prices, k) {
+    const firstPrice = prices[0];
+    const dp = Array(k + 1).fill(null).map(() => ({
+        maxProfit: 0,
+        buyHold: -firstPrice,
+        sellHold: firstPrice
+    }));
+    for(let i = 0;i < prices.length;++i) {
+      const current = prices[i];
+      for(let j = k;j > 0;--j) {
+          const prevProfit = dp[j - 1].maxProfit;
+          dp[j].maxProfit = Math.max(dp[j].maxProfit, dp[j].buyHold + current, dp[j].sellHold - current);
+          dp[j].buyHold = Math.max(dp[j].buyHold, prevProfit - current);
+          dp[j].sellHold = Math.max(dp[j].sellHold, prevProfit + current);
+      }
+    }
+    return dp[k].maxProfit;
+};
+let prices = [1,7,9,8,2], k = 2;
+// 14
+// can make $14 of profit through 2 transactions:
+// A normal transaction: buy the stock on day 0 for $1 then sell it on day 2 for $9. 
+// A short selling transaction: sell the stock on day 3 for $8 then buy back on day 4 for $2.
+// 9 - 1 = 8 ; 8 - 2 = 6; 8 + 6 = 14
+// console.log(maximumProfit(prices,k))

javascript/LeetCode/String/1859.js

@@ -0,0 +1,28 @@
+/**
+ * 1859. Sorting the Sentence
+ * 
+ * 字串句子中每個單字最後面都會有個數字，利用該數字重新將句子做排序，但結果每個單字不能有數字。
+ * 
+ * @param {string} s
+ * @return {string}
+ */
+var sortSentence = function(s) {
+    let res = "";
+    let splitS = s.split(" ");
+    let map = new Map();
+    for(let i = 0;i < splitS.length;++i){
+      // 取得每個字的最後一個數字，並把它放入new map => key = int,value = str
+      const lastInt = parseInt(splitS[i].substring(splitS[i].length - 1));
+      map.set(lastInt,splitS[i].substring(0,splitS[i].length - 1));
+    }
+    // sort map
+    var mapAsc = new Map([...map.entries()].sort());
+    for(const [key,value] of mapAsc) {
+      res += " "+value;
+    }
+    return res.trim()
+};
+let s = "is2 sentence4 This1 a3";
+// Output: "This is a sentence"
+// Explanation: Sort the words in s to their original positions "This1 is2 a3 sentence4", then remove the numbers.
+console.log(sortSentence(s));

javascript/index.js

@@ -1176,40 +1176,5 @@ let nums = [8,4,2,8,4];
 // console.log(specialTriplets(nums));
 
 
-/**
- * 3573. Best Time to Buy and Sell Stock V
- * 
- * 正常交易：i天買，j天賣(i < j)，利潤：proces[j] - prices[i]
- * 炒短線：i天賣，j天買回來(i < j)，利潤：prices[i] - prices[j]
- * 不能同一天買賣單張股票
- * 回傳在交易k次後最多能賺多少
- * 
- * @param {number[]} prices 第i天的股價
- * @param {number} k 最多只能交易k次
- * @return {number}
- */
-var maximumProfit = function(prices, k) {
-    const firstPrice = prices[0];
-    const dp = Array(k + 1).fill(null).map(() => ({
-        maxProfit: 0,
-        buyHold: -firstPrice,
-        sellHold: firstPrice
-    }));
-    for(let i = 0;i < prices.length;++i) {
-      const current = prices[i];
-      for(let j = k;j > 0;--j) {
-          const prevProfit = dp[j - 1].maxProfit;
-          dp[j].maxProfit = Math.max(dp[j].maxProfit, dp[j].buyHold + current, dp[j].sellHold - current);
-          dp[j].buyHold = Math.max(dp[j].buyHold, prevProfit - current);
-          dp[j].sellHold = Math.max(dp[j].sellHold, prevProfit + current);
-      }
-    }
-    return dp[k].maxProfit;
-};
-let prices = [1,7,9,8,2], k = 2;
-// 14
-// can make $14 of profit through 2 transactions:
-// A normal transaction: buy the stock on day 0 for $1 then sell it on day 2 for $9. 
-// A short selling transaction: sell the stock on day 3 for $8 then buy back on day 4 for $2.
-// 9 - 1 = 8 ; 8 - 2 = 6; 8 + 6 = 14
-console.log(maximumProfit(prices,k))
+
+