Day 32 | 122. Best Time to Buy and Sell Stock II | 55. Jump Game | 45. Jump Game II

原创已于 2023-01-17 14:12:57 修改 · 197 阅读

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于 2023-01-15 13:49:34 首次发布

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该文记录了LeetCode中一系列与数组、链表相关的编程挑战，包括买卖股票的最佳时机、跳跃游戏、最大跳跃数等，展示了如何运用算法解决这些经典问题。

LeetCode 122. Best Time to Buy and Sell Stock II

Question Link

class Solution {
    public int maxProfit(int[] prices) {
        int result = 0;
        for(int i = 1; i < prices.length; i++){
            int profit = prices[i] - prices[i - 1];
            if(profit > 0)
                result += profit;
        }
        return result;
    }
}

We can decompose the final profit by day. The max profit is the sum of the profit of days whose profit is positive.

LeetCode 55. Jump Game

Question Link

class Solution {
    public boolean canJump(int[] nums) {
        int coverRange = 0;
        for(int i = 0; i <= coverRange; i++){
            coverRange = Math.max(coverRange, i + nums[i]);
            if(coverRange >= nums.length - 1)
                return true;
        }

        return false;
    }
}

The initial cover range should be 0
We only can update the cover range when i in the cover range

LeetCode 45. Jump Game II

Question Link

class Solution {
    public int jump(int[] nums) {
        if(nums == null || nums.length == 0 || nums.length == 1)
            return 0;
        // The number of jumps
        int count = 0;
        // The current cover range
        int curDistance = 0;
        // The max cover range
        int maxDistance = 0;
        for(int i = 0; i < nums.length; i++){
            maxDistance = Math.max(maxDistance, i + nums[i]);
            // Jump one more step can reach the end position.
            if(maxDistance >= nums.length - 1){
                count++;
                break;
            }
            // Update the distance that the next step can reach when reaching the current max cover range
            if(curDistance == i){
                count++;
                curDistance = maxDistance;
            }
        }
        return count;
    }
}