链接: https://leetcode-cn.com/problems/find-good-days-to-rob-the-bank/
思路:
前缀和: g数组:表示下标i与i - 1 的大小关系 规定:security[i] > security[i - 1] g[i] = 1; security[i] < security[i - 1] g[i] = -1; security[i] = security[i - 1] g[i] = 0; 在 [i - time, i + time) 内满足: (i - time, i] 1的个数为0 (i, i + time] -1的个数为0 需要左开右闭区间,使用前缀和时,用i + 1位置记录i的状态;a[i]表示i 与 i - 1关系,但是此时i 和 i - 1的关系并不关心
代码:
public List<Integer> goodDaysToRobBank(int[] security, int time) {
if (security.length < 2 * time + 1) return new ArrayList<>();
List<Integer> res = new ArrayList<>();
int len = security.length;
int a[] = new int[len + 1];
int b[] = new int[len + 1];
int g[] = new int[len];
for (int i = 1; i < len; i++) {
if (security[i] == security[i - 1]) continue;
g[i] = security[i] < security[i - 1] ? -1 : 1;
}
for (int i = 1; i <= len ;i++) a[i] = a[i - 1] + (g[i - 1] == 1 ? 1 : 0);
for (int i = 1; i <= len ;i++) b[i] = b[i - 1] + (g[i - 1] == -1 ? 1 : 0);
for (int i = time; i < len - time; i++) {
if (a[i + 1] - a[i - time + 1] > 0) continue;
if (b[i + 1 + time] - b[i + 1] > 0) continue;
res.add(i);
}
return res;
}