上岸算法 I LeetCode Weekly Contest 232解题报告

No.1 仅执行一次字符串交换能否使两个字符串相等

解题思路

签到题，枚举一遍统计出所有不想等的位置。

代码展示

class Solution {
    public boolean areAlmostEqual(String s1, String s2) {
        List<Integer> differ = new ArrayList<>();
        for (int i = 0; i < s1.length(); i++) {
            if (s1.charAt(i) != s2.charAt(i)) {
                differ.add(i);
            }
        }
        return differ.size() == 0 ||
                (differ.size() == 2 &&
                        s1.charAt(differ.get(0)) == s2.charAt(differ.get(1)) &&
                        s1.charAt(differ.get(1)) == s2.charAt(differ.get(0))
                );
    }
}

No.2 找出星型图的中心节点

解题思路

n 个点 n - 1 条边必然是树。因为是星型图，所以叶子结点的度数均为 1. 只要一个节点在边中出现了两次以上，它必然是中心节点。

代码展示

class Solution {
    public int findCenter(int[][] edges) {
        Set<Integer> set = new HashSet<>();
        for (var edge : edges) {
            for (int e : edge) {
                if (set.contains(e)) {
                    return e;
                }
                set.add(e);
            }
        }
        return -1;
    }
}

No.3 最大平均通过率

解题思路

贪心，每次放一个学生进入某一个班级，进入能提升通过率提升地最多的那个班级。

代码展示

class Solution {
    static class Class {
        int pass;
        int total;

        public Class(int pass, int total) {
            this.pass = pass;
            this.total = total;
        }

        double differ() {
            return (double) (pass + 1) / (total + 1) - (double) pass / total;
        }
    }

    public double maxAverageRatio(int[][] classes, int extraStudents) {
        PriorityQueue<Class> heap = new PriorityQueue<>((c1, c2) -> c1.differ() > c2.differ() ? -1 : 1);
        for (var c : classes) {
            heap.add(new Class(c[0], c[1]));
        }
        for (int i = 0; i < extraStudents; i++) {
            Class c = heap.poll();
            heap.add(new Class(c.pass + 1, c.total + 1));
        }
        double sum = 0;
        while (!heap.isEmpty()) {
            Class c = heap.poll();
            sum += (double) c.pass / c.total;
        }
        return sum / classes.length;
    }
}

No.4 好子数组的最大分数

解题思路

单调栈，与最大柱状图类似，找到一个值的左右侧第一个比它小的即可。

代码展示

class Solution {
    public int maximumScore(int[] nums, int k) {
        int[] heap = new int[nums.length + 5];
        int top = 0;
        int res = 0;
        for (int i = 0; i < nums.length; i++) {
            while (top > 0 && nums[heap[top]] > nums[i]) {
                int right = i;
                int left = top > 1 ? heap[top - 1] : -1;
                if (left < k && right > k) {
                    res = Math.max(res, (right - left - 1) * nums[heap[top]]);
                }
                top--;
            }
            heap[++top] = i;
        }
        while (top > 0) {
            int right = nums.length;
            int left = top > 1 ? heap[top - 1] : -1;
            if (left < k && right > k) {
                res = Math.max(res, (right - left - 1) * nums[heap[top]]);
            }
            top--;
        }
        return res;
    }
}