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