lcp341

一最多的行

方法一：模拟

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class Solution {
    public int[] rowAndMaximumOnes(int[][] mat) {
        int m = mat.length, n = mat[0].length, max = 0;
        int[] res = new int[2];
        for (int i = 0; i < m; ++i) {
            int count = 0;
            for (int j = 0; j < n; ++j) {
                if (mat[i][j] == 1)
                    ++count;
            }
            if (count > max) {
                res[0] = i;
                res[1] = count;
                max = count;
            }
        }
        return res;
    }
}

找出可整除性得分最大的整数

方法一：整除与被整除？

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class Solution {
    public int maxDivScore(int[] nums, int[] divisors) {
        int max = 0, val = divisors[0];
        Set<Integer> set = new HashSet<>();
        for (int i = 0; i < divisors.length; ++i) {
            if (set.contains(divisors[i]))
                continue;
            set.add(divisors[i]);
            int count = 0;
            for (int j = 0; j < nums.length; ++j) {
                int num = nums[j];
                if (nums[j] % divisors[i] == 0)
                    ++count;
            }
            if (count > max || (count == max && divisors[i] < val)) {
                max = count;
                val = divisors[i];
            }
        }
        return val;
    }
}

构造有效字符串的最少插入数

方法一：bullshit

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class Solution {
    public int addMinimum(String word) {
        if (word.equals("abc"))
            return 0;
        int res = 0, n = word.length();
        int flag = 1;
        for (int i = 0; i < n; ++i) {
            char ch = word.charAt(i);
            if (ch == 'a') {
                if (i + 2 < n && word.charAt(i + 1) == 'b' && word.charAt(i + 2) == 'c')
                    i = i + 2;
                else if (i + 2 >= n) {
                    if (i == n - 1)
                        res += 2;
                    else if (i == n - 2 && word.charAt(i + 1) != 'a') {
                        res += 1;
                        i = n - 1;
                    }
                    else if (i == n - 2 && word.charAt(i + 1) == 'a')
                        res += 2;
                }
                else if (word.charAt(i + 1) == 'a') {
                    res += 2;
                }
                else if (word.charAt(i + 1) != 'a') {
                    res += 1;
                    i = i + 1;
                }
                else if (word.charAt(i + 2) != 'c') {
                    res += 1;
                    i += 1;
                }
            } else if (ch == 'b') {
                if (i + 1 < n) {
                    if (word.charAt(i + 1) == 'c') {
                        res += 1;
                        ++i;
                    }
                    else {
                        res += 2;
                    }
                }
                else if (i == n - 1) {
                    res += 2;
                }
            }
            else  { // 'c'
                res += 2;
            }
        }
        return res;
    }
}

最小化旅行的价格总和

Contest > LCP
#力扣杯
lcp341
https://leopol1d.github.io/2023/08/12/lcp341/
作者
Leopold
发布于
2023年8月12日
许可协议