Grid Illumination

2019-07-07 16:53:31

问题描述:

问题求解:

本题和n后问题很类似,所以最初的时候就直接套了n后的板子,MLE。

 public int[] gridIllumination(int N, int[][] lamps, int[][] queries) {
        int[] rows = new int[N];
        int[] cols = new int[N];
        int[] diag1 = new int[2 * N - 1];
        int[] diag2 = new int[2 * N - 1];
        Set<String> set = new HashSet<>();
        for (int[] lamp : lamps) {
            int x = lamp[0];
            int y = lamp[1];
            rows[x]++;
            cols[y]++;
            diag1[x + y]++;
            diag2[N - 1 - x + y]++;
            set.add(String.valueOf(x) + "_" + String.valueOf(y));
        }
        int n = queries.length;
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            int[] q = queries[i];
            int x = q[0];
            int y = q[1];
            if (rows[x] > 0 || cols[y] > 0 || diag1[x + y] > 0 || diag2[N - 1 - x + y] > 0) 
                res[i] = 1;
            else res[i] = 0;
            helper(x, y, set, rows, cols, diag1, diag2, N);
        }
        return res;
    }
    
    private void helper(int x, int y, Set<String> set, int[] rows, int[] cols, int[] diag1, int[] diag2, int N) {
        for (int i = -1; i <= 1; i++) {
            for (int j = -1; j <= 1; j++) {
                int pi = x + i;
                int pj = y + j;
                if (pi < 0 || pi >= N || pj < 0 || pj >= N) continue;
                String cur = String.valueOf(pi) + "_" + String.valueOf(pj);
                if (set.contains(cur)) {
                    set.remove(cur);
                    rows[pi]--;
                    cols[pj]--;
                    diag1[pi + pj]--;
                    diag2[N - 1 - pi + pj]--;
                }
            }
        }
    }

那么本题的核心就是如何降低空间复杂度了,如何做呢?

降低空间复杂度有两种常用的技巧:

1. 将数组转为HashMap,这样就可以在大量稀疏的数组中提取到有用的信息,避免无用的信息。

2. 如何保存已经点亮的位置,这里采用字符串的处理显然是低效的,最好的方法是采用位操作的方式,使用long将两个int拼接起来达到区分的效果。

    public int[] gridIllumination(int N, int[][] lamps, int[][] queries) {
        HashMap<Integer, Integer> rows, cols, diag1, diag2;
        rows = new HashMap<Integer, Integer>();
        cols = new HashMap<Integer, Integer>();
        diag1 = new HashMap<Integer,Integer>();
        diag2 = new HashMap<Integer, Integer>();
        Set<Long> set = new HashSet<>();
        for (int[] lamp : lamps) {
            int x = lamp[0];
            int y = lamp[1];
            rows.put(x, rows.getOrDefault(x, 0) + 1);
            cols.put(y, cols.getOrDefault(y, 0) + 1);
            diag1.put(x + y, diag1.getOrDefault(x + y, 0) + 1);
            diag2.put(N - 1 - x + y, diag2.getOrDefault(N - 1 - x + y, 0) + 1);
            set.add((long)x << 32 | (long)y);
        }
        int n = queries.length;
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            int[] q = queries[i];
            int x = q[0];
            int y = q[1];
            if (rows.getOrDefault(x, 0) > 0 || cols.getOrDefault(y, 0) > 0 || diag1.getOrDefault(x + y, 0) > 0 || diag2.getOrDefault(N - 1 - x + y, 0) > 0) 
                res[i] = 1;
            else res[i] = 0;
            helper(x, y, set, rows, cols, diag1, diag2, N);
        }
        return res;
    }
    
    private void helper(int x, int y, Set<Long> set,  HashMap<Integer, Integer> rows, HashMap<Integer, Integer> cols, HashMap<Integer, Integer> diag1, HashMap<Integer, Integer> diag2, int N) {
        for (int i = -1; i <= 1; i++) {
            for (int j = -1; j <= 1; j++) {
                int pi = x + i;
                int pj = y + j;
                if (pi < 0 || pi >= N || pj < 0 || pj >= N) continue;
                long cur = (long)pi << 32 | pj;
                if (set.contains(cur)) {
                    set.remove(cur);
                    rows.put(pi, rows.get(pi) - 1);
                    cols.put(pj, cols.get(pj) - 1);
                    diag1.put(pi + pj, diag1.get(pi + pj) - 1);
                    diag2.put(N - 1 - pi + pj, diag2.get(N - 1 - pi + pj) - 1);
                }
            }
        }
    }

  

原文地址:https://www.cnblogs.com/hyserendipity/p/11146784.html