leetcode 中等难度

L494

//方法一：回溯法
class Solution {
    public static int findTargetSumWays(int[] nums, int target) {
        cnt =0;
        recurve(nums, 0, 0, target);
        return cnt;
    }

    static int cnt = 0;

    public static void recurve(int[] nums, int n, int sum, int target) {
        if (n == nums.length) {
            if (sum == target) cnt++;
            return;
        }
        recurve(nums, n + 1, sum + nums[n], target);
        recurve(nums, n + 1, sum - nums[n], target);
    }
}
//方法二：深度优先搜索

406. 根据身高重建队列

方法一：堆排序
public static int[][] reconstructQueue(int[][] people) {
        PriorityQueue<Integer[]> heap = sort(people);
        LinkedList<Integer[]> result = new LinkedList<>();
        //插入排序
        for (int i = 0; i < people.length; i++) {
            Integer[] x = heap.poll();
            result.add(x[1], x);
        }
        int[][] res = new int[people.length][2];
        for (int i = 0; i < people.length; i++) {
            Integer[] t = result.get(i);
            res[i][0] = t[0];
            res[i][1] = t[1];
        }
        return res;
    }


    public static PriorityQueue<Integer[]> sort(int[][] people) {
        PriorityQueue<Integer[]> heap = new PriorityQueue<>(new Comparator<Integer[]>() {
            @Override
            public int compare(Integer[] o1, Integer[] o2) {
                if (o1 == null || o2 == null) return 0;
                if (o1[0].equals(o2[0])) {
                    return o1[1] - o2[1];
                }
                return o2[0] - o1[0];
            }
        });
        for (int i = 0; i < people.length; i++) {
            heap.add(new Integer[]{people[i][0], people[i][1]});
        }
        return heap;
    }
方法二：

leetcode 完全平方数

//方法1：