每日一题

954. 二倍数对数组(抄)

抽空好好看看。里面的语法我都看不懂。

class Solution {
    public boolean canReorderDoubled(int[] arr) {
        Map<Integer, Integer> cnts = new HashMap<>();
        for(int num: arr)
            cnts.put(num, cnts.getOrDefault(num, 0) + 1);
        List<Integer> list = new ArrayList<>(cnts.keySet());
        Collections.sort(list, (a, b) -> a - b);
        for(int key: list) {
            if(key > 0) {
                if(cnts.getOrDefault(key * 2, 0) < cnts.get(key))
                    return false;
                if(cnts.get(key) > 0)
                    cnts.put(key * 2, cnts.get(key * 2) - cnts.get(key));
            } else if(key == 0) {
                if(cnts.get(key) % 2 != 0)
                    return false;
            }
            else {
                if(cnts.get(key) > 0 && (key % 2 != 0 || cnts.getOrDefault(key/2, 0) < cnts.get(key)))
                    return false;
                if(cnts.get(key) > 0)
                    cnts.put(key / 2, cnts.get(key / 2) - cnts.get(key));
            }
        }
        return true;
    }
}

169. 多数元素(抄)

其实我并不是很理解。

class Solution {
    public int majorityElement(int[] nums) {
        Arrays.sort(nums);
        return nums[nums.length >> 1];
    }
}

剑指 Offer II 100. 三角形中最小路径之和

class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int n = triangle.size();
        int[][] f = new int[2][n];
        f[0][0] = triangle.get(0).get(0);
        for (int i = 1; i < n; ++i) {
            int curr = i % 2;
            int prev = 1 - curr;
            f[curr][0] = f[prev][0] + triangle.get(i).get(0);
            for (int j = 1; j < i; ++j) {
                f[curr][j] = Math.min(f[prev][j - 1], f[prev][j]) + triangle.get(i).get(j);
            }
            f[curr][i] = f[prev][i - 1] + triangle.get(i).get(i);
        }
        int minTotal = f[(n - 1) % 2][0];
        for (int i = 1; i < n; ++i) {
            minTotal = Math.min(minTotal, f[(n - 1) % 2][i]);
        }
        return minTotal;
    }
}