一个字符串s，表示仓库的墙 与 货物，其中’|‘表示墙,’'表示货物。
给定一个起始下标start和一个终止下标end，
找出子串中 被墙包裹的货物 数量。
比如：
s = "|||"，
start = 1, end = 7，
start和end截出的子串是 "||"，
被 '|'包裹的 '’ 有两个，所以返回2，
现在给定一系列的start，startIndices[]，和对应一系列的end ,endIndices[]。
返回每一对[start,end]的截出来的货物数量。
数据规模：
字符串s长度<=10^5，
startIndices长度 == endIndices长度 <=10^5。
亚马逊的货物和墙的问题。
前缀和。
时间复杂度：O(N)。
空间复杂度：O(N)。
代码用rust编写。代码如下：
fn main() {
let s = "|**|**|*";
let mut a = vec![0, 1, 3, 4];
let mut b = vec![7, 7, 6, 5];
let ans = number(s, &mut a, &mut b);
println!("ans = {:?}", ans);
}
fn number(s: &str, starts: &mut Vec<i32>, ends: &mut Vec<i32>) -> Vec<i32> {
let str = s.as_bytes();
let n = str.len() as i32;
let mut left: Vec<i32> = vec![];
let mut right: Vec<i32> = vec![];
let mut sum: Vec<i32> = vec![];
for _ in 0..n {
left.push(0);
right.push(0);
sum.push(0);
}
let mut pre = -1;
let mut num = 0;
for i in 0..n {
pre = if str[i as usize] == '|' as u8 { i } else { pre };
num += if str[i as usize] == '*' as u8 { 1 } else { 0 };
left[i as usize] = pre;
sum[i as usize] = num;
}
pre = -1;
let mut i = n - 1;
while i >= 0 {
pre = if str[i as usize] == '|' as u8 { i } else { pre };
right[i as usize] = pre;
i -= 1;
}
let m = starts.len() as i32;
let mut ans: Vec<i32> = vec![];
for _ in 0..m {
ans.push(0);
}
for i in 0..m {
ans[i as usize] = stars(
starts[i as usize],
ends[i as usize],
&mut left,
&mut right,
&mut sum,
);
}
return ans;
}
fn stars(start: i32, end: i32, l: &mut Vec<i32>, r: &mut Vec<i32>, s: &mut Vec<i32>) -> i32 {
let left = r[start as usize];
let right = l[end as usize];
if left == -1 || right == -1 || (left >= right) {
return 0;
}
return if left == 0 {
s[right as usize]
} else {
s[right as usize] - s[(left - 1) as usize]
};
}执行结果如下：
