Leetcode 佛系刷題 每天動動腦

• https://leetcode.com/problemset/all/
• 沒有特殊標明的話，貼上來的代碼都是在leetcode上跑成功，速度100%的。

11. Container With Most Water

Given n non-negative integers a1, a2, …, an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
Example:
Input: [1,8,6,2,5,4,8,3,7]
Output: 49

• 分治，剪枝

    public int maxArea(int[] height) {
        int max = 0;
        int startHeight = 0;
        for (int i = 0; i < height.length - 1; i++) {
            if (height[i] <= startHeight || (height.length - i) * height[i] < max) {
                continue;
            }
            int cur = maxSingleArea(i, height);
            if (cur > max) {
                max = cur;
                startHeight = height[i];
            }
        }
        return max;
    }

    public int maxSingleArea(int i, int[] height) {
        int max = 0;
        for (int j = height.length - 1; j > i; j--) {
            int cur = 0;
            if (height[j] >= height[i]) {
                cur = (j - i) * height[i];
                max = cur > max ? cur : max;
                return max;
            } else {
                cur = (j - i) * height[j];
                max = cur > max ? cur : max;
            }
        }
        return max;
    }

• 貪心算法，前後逐漸向內收縮

public int maxArea(int[] height) {
        int max = 0;
        int startHeight = 0;
        int endHeight = 0;
        int i = 0;
        int j = height.length - 1;
        while (i < j) {
            if (height[i] < startHeight) {
                i++;
                continue;
            }
            if (height[j] < endHeight) {
                j--;
                continue;
            }
            int cur = 0;
            if (height[j] >= height[i]) {
                cur = (j - i) * height[i];
                if (cur > max) {
                    max = cur;
                    endHeight = height[j];
                }
                i++;
            } else {
                cur = (j - i) * height[j];
                if (cur > max) {
                    max = cur;
                    startHeight = height[i];
                }
                j--;
            }
        }

        return max;
    }

7. Reserver Integer

Given a 32-bit signed integer, reverse digits of an integer.

Example	Input	Output
1	123	321
2	-123	-321
3	120	21

• Note: Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

    public int reverse(int x) {
        boolean negative = false;
        if (x < 0) {
            negative = true;
            x = -x;
        }
        long result = 0L;
        while (x > 0) {
            result *= 10;
            result += x % 10;
            x /= 10;
        }
        if (result > Integer.MAX_VALUE) {
            return 0;
        }
        if (negative) {
            result = -result;
        }
        return (int) result;
    }