标签：else span turn contains csharp code 信息 hashset 拼接
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);
}
}
}
}
标签：else span turn contains csharp code 信息 hashset 拼接
原文地址：https://www.cnblogs.com/TIMHY/p/11146784.html
