2100. 适合打劫银行的日子

题目：
leetcode.cn/problems/fi…
算法：
方法一：前缀和

func goodDaysToRobBank(security []int, time int) []int {
    // 初始化数组g, g[i] = 0， security[i-1]=security[i]
    // g[i] = 1， security[i-1] < security[i]
    // g[i] = -1， security[i-1] > security[i]
    // 如果i天前连续time天警卫数非递增，则有g[i] - t[i - time - 1] == time
    ans := make([]int, 0)
    n := len(security)
    g := make([]int, n)
    for i := 1; i < n; i ++ {
        if security[i] > security[i - 1] {
            g[i] = 1
        } else if security[i] < security[i - 1] {
            g[i] = -1
        }
    }
    // a[i + 1], b[i + 1]分别代表security[i]之前大于等于security[i]，之前小于等于security[i]的数量
    a, b := make([]int, n + 1), make([]int, n + 1)
    for i := range g {
        if g[i] == 0 {
            a[i + 1] = a[i]
            b[i + 1] = b[i]
        } else if g[i] == 1 {
            a[i + 1] = a[i] + 1
            b[i + 1] = b[i]
        } else {
            a[i + 1] = a[i]
            b[i + 1] = b[i] + 1
        }
    }
    for i := time; i + time < n; i ++ {
        descAndEqCount := a[i + 1] - a[i + 1 - time ]
        ascAndEqCount :=  b[i + time + 1] - b[i + 1]
        if descAndEqCount == 0 && ascAndEqCount == 0 {
            ans = append(ans, i)
        }
    }
    return ans
}