Merge pull request #151 from clingoram/mavis

新增解題練習

Author: clingoram

javascript/LeetCode/Array/1184.js

@@ -0,0 +1,33 @@
+/**
+ * 1184. Distance Between Bus Stops
+ * 
+ * 從圓中(distance)找出從start到destination最短的距離
+ * 
+ * @param {number[]} distance
+ * @param {number} start
+ * @param {number} destination
+ * @return {number}
+ */
+var distanceBetweenBusStops = function(distance, start, destination) {
+  let res = 0,total = distance.reduce((a,b)=>a + b);
+  if(start > destination){
+    let temp = start;
+    start = destination;
+    destination = temp;
+  }
+  for(let i = 0;i < distance.length;++i) {
+    if(i >= start && i < destination){
+      res += distance[i]
+    }
+  }
+  return Math.min(res,total - res);
+};
+//             [0,1,2,3]
+// let distance = [1,2,3,4], start = 0, destination = 2;
+// ans.3 => 2+1
+// 7 => 1+2+4
+// Distance between 0 and 2 is 3 or 7, minimum is 3.
+
+let distance = [7,10,1,12,11,14,5,0], start = 7, destination = 2;
+// 17
+console.log(distanceBetweenBusStops(distance,start,destination));

javascript/LeetCode/Array/169.js

@@ -0,0 +1,23 @@
+/**
+ * 169. Majority Element
+ *
+ * 找出majority element  = 出現次數大於 nums.length / 2
+ * @param {number[]} nums
+ * @return {number}
+ */
+var majorityElement = function(nums) {
+    let map = new Map();
+    let ans = 0;
+    for(const i of nums){
+      map.has(i) ? map.set(i,map.get(i)+1) : map.set(i,1);
+    }
+    for(const [key,value] of map) {
+      if(value > (nums.length / 2)){
+        ans = key;
+      }
+    }
+    return ans;
+};
+let nums = [2,2,1,1,1,2,2];
+// 2
+console.log(majorityElement(nums));

javascript/LeetCode/Array/1779.js

@@ -0,0 +1,43 @@
+/**
+ * 1779. Find Nearest Point That Has the Same X or Y Coordinate
+ * 
+ * x & y = current location
+ * A point is valid if it shares the same x-coordinate or the same y-coordinate as your location.
+ * 傳回與目前位置曼哈頓距離最小的有效點的索引（從 0 開始索引）。
+ * 如果存在多個有效點，則傳回索引最小的有效點。如果沒有有效點，則傳回 -1。
+ * 
+ * The Manhattan distance between two points (x1, y1) and (x2, y2) is abs(x1 - x2) + abs(y1 - y2).
+ * 
+ * @param {number} x
+ * @param {number} y
+ * @param {number[][]} points
+ * @return {number}
+ */
+var nearestValidPoint = function(x, y, points) {
+  let ans = -1;
+  let smallest = Infinity;
+  for(let i = 0;i < points.length;++i) {
+    let prev = Math.abs(x - parseInt(points[i][0]));
+    let next = Math.abs(y - parseInt(points[i][1]));
+    if(prev * next === 0 && (prev + next) < smallest){
+      smallest = prev + next;
+      ans = i;
+    }
+    
+  }
+  return ans;
+};
+// let x = 3, y = 4, points = [[1,2],[3,1],[2,4],[2,3],[4,4]];
+// 2
+// Of all the points, only [3,1], [2,4] and [4,4] are valid. 
+// Of the valid points, [2,4] and [4,4] have the smallest Manhattan distance from your current location, with a distance of 1. 
+// [2,4] has the smallest index, so return 2.
+
+// let x = 3, y = 4, points = [[3,4]];
+// 0
+// let x = 3, y = 4, points = [[2,3]];
+// -1
+
+let x = 5, y = 1, points = [[1,1],[6,2],[1,5],[3,1]];
+// 3
+console.log(nearestValidPoint(x,y,points));

javascript/LeetCode/Array/2154.js

@@ -0,0 +1,35 @@
+/**
+ * 2154. Keep Multiplying Found Values by Two
+ * 
+ * 初次先尋找original是否在nums內，若不在直接回傳original
+ * 若在，則original * 2 = 新的original （重複此步驟直到找不到original為止）
+ * 
+ * @param {number[]} nums
+ * @param {number} original
+ * @return {number}
+ */
+var findFinalValue = function(nums, original) {
+    nums.sort((a,b) => a - b);
+    for(let i = 0;i < nums.length;++i) {
+      if(nums.includes(original)){
+        original *= 2;
+      }
+    }
+    return original;
+
+    // solution 2
+    // nums.sort((a,b) => a - b);
+    // for(const a of nums){
+    //   if(a === original){
+    //     original*=2;
+    //   }
+    // }
+    // return original;
+};
+// let nums = [5,3,6,1,12], original = 3;
+// 24
+// let nums = [2,7,9], original = 4;
+// 4
+let nums = [8,19,4,2,15,3], original = 2;
+// 16
+console.log(findFinalValue(nums,original));

javascript/LeetCode/Array/219.js

@@ -76,6 +76,18 @@ var containsNearbyDuplicate = function (nums, k) {
         toMap.set(nums[i], i);
     }
     return false;
+
+    // solution 3.
+    // i & j => nums[i] == nums[j] & abs(i - j) <= k.
+    // nums[i]跟nums[j]一樣
+    // let map = new Map();
+    // for(let i = 0;i < nums.length;++i){
+    //   if(map.has(nums[i]) && i - map.get(nums[i]) <= k){
+    //     return true
+    //   }
+    //   map.set(nums[i],i);
+    // }
+    // return false;
 };
 const nums = [1, 2, 3, 1], k = 3; // return true
 //Input: nums = [1,0,1,1], k = 1 return true

javascript/LeetCode/Array/3151.js

@@ -0,0 +1,27 @@
+/**
+ * 3151. Special Array I
+ * 
+ * 檢查陣列任兩個相鄰元素的奇偶性是否都不同。若是回傳true
+ * @param {number[]} nums
+ * @return {boolean}
+ */
+var isArraySpecial = function(nums) {
+  if(nums.length === 1){
+    return true;
+  }
+  for(let i = 0;i < nums.length;++i) {
+    let previous = nums[i];
+    let next = nums[i + 1];
+    if(previous % 2 !== next % 2){
+      continue;
+    }else{
+      return false;
+    }
+  }
+  return true;
+};
+// let nums = [4,3,1,6];
+// false
+let nums = [2,1];
+// true
+console.log(isArraySpecial(nums))

javascript/LeetCode/Array/3190.js

@@ -0,0 +1,20 @@
+/**
+ * 3190. Find Minimum Operations to Make All Elements Divisible by Three
+ * 
+ * 計算能將所有元素被3整除的步驟為幾，一次操作能將元素+1或-1。
+ * 
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minimumOperations = function(nums) {
+  let answer = 0;
+  for(let i = 0;i < nums.length;++i) {
+    if(nums[i] % 3 !== 0){
+      answer++;
+    }
+  }
+  return answer;
+};
+let nums = [1,2,3,4];
+// 3
+console.log(minimumOperations(nums));

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/Array/3731.js

@@ -0,0 +1,20 @@
+/**
+ * 3731. Find Missing Elements
+ * 
+ * 找出參數陣列中缺少的數字，並以陣列型態回傳
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var findMissingElements = function(nums) {
+  let ans = [];
+  let min = Math.min(...nums),max = Math.max(...nums);
+  for(let i = min;i <= max;++i){
+    if(nums.includes(i) === false){
+      ans.push(i);
+    }
+  }
+  return ans;
+};
+let nums = [5,1];
+// [2,3,4]
+console.log(findMissingElements(nums));

javascript/LeetCode/String/168.js

@@ -0,0 +1,20 @@
+/**
+ * 168. Excel Sheet Column Title
+ * 
+ * 給定一個整數 columnNumber，傳回其在 Excel 表格中對應的列標題。
+ * @param {number} columnNumber
+ * @return {string}
+ */
+var convertToTitle = function(columnNumber) {
+    let ans = "";
+    let baseASCIICode = "A".charCodeAt();
+    while(columnNumber > 0){
+      let offset = (columnNumber - 1) % 26;
+      ans = String.fromCharCode(baseASCIICode + offset) + ans;
+      columnNumber = parseInt((parseInt(columnNumber - 1) / 26));
+    }
+    return ans;
+};
+let columnNumber = 701;
+// "ZY";
+console.log(convertToTitle(columnNumber));