LeetCode算法题---第3天

注:大佬解答来自LeetCode官方题解

121.买卖股票的最佳时期

1.题目

2.个人解答

function maxProfit(prices) {//更新最低价格和最大利润let minPrice = prices[0];let maxProfit = 0;for (let i = 1; i < prices.length; i++) {// 如果当前价格比最低价格还低，更新最低价格if (prices[i] < minPrice) {minPrice = prices[i];}// 计算当前价格卖出时的利润，并更新最大利润else if (prices[i] - minPrice > maxProfit) {maxProfit = prices[i] - minPrice;}}return maxProfit;
}

 3.大佬解答

122.买卖股票最佳时期Ⅱ

1.题目

2.个人解答

var maxProfit = function (prices) {//更新最低价格和最大利润let minPrice = prices[0];let maxProfit = 0;for (let index = 1; index < prices.length; index++) {if (prices[index] < minPrice) {minPrice = prices[index];} else {maxProfit += prices[index]-minPrice;minPrice=prices[index]}}return maxProfit
};

 3.大佬解答

var maxProfit = function(prices) {const n = prices.length;let dp0 = 0, dp1 = -prices[0];for (let i = 1; i < n; ++i) {let newDp0 = Math.max(dp0, dp1 + prices[i]);let newDp1 = Math.max(dp1, dp0 - prices[i]);dp0 = newDp0;dp1 = newDp1;}return dp0;
};

 

 

var maxProfit = function(prices) {let ans = 0;let n = prices.length;for (let i = 1; i < n; ++i) {ans += Math.max(0, prices[i] - prices[i - 1]);}return ans;
};

 55.跳跃游戏

1.题目

2.个人解答

function canJump(nums) {let maxReach = 0;for (let i = 0; i < nums.length; i++) {if (i > maxReach) {return false; // 如果当前位置无法到达，则返回false}maxReach = Math.max(maxReach, i + nums[i]); // 更新maxReach}return maxReach >= nums.length - 1;
}

 3.大佬解答