LeetCode —— Trapping Rain Water

链接：http://leetcode.com/onlinejudge#question_42

原题：

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap
after raining.

For example,

Given

[0,1,0,2,1,0,1,3,2,1,2,1]

,
return

6

.



The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks
Marcos for contributing this image!
思路：

我是先将数组排序，从大到小。用左右两个left，right下标，从最大值开始，先两边扩散，

如果落在[left，right]之间的，就pass。

这样的复杂度是O（nlogn）的，主要是排序上面。

不过discussion里面有一个强大的O（n）方法。

代码：

bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
return a.first > b.first;
}

class Solution {
public:
int trap(int A[], int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (n <= 2)
return 0;

vector<pair<int, int> > vec;
for (int i=0; i<n; i++)
vec.push_back(pair<int, int>(A[i], i));

sort(vec.begin(), vec.end(), cmp);

int total = 0;
int left, right;
left = right = vec[0].second;
int count = 1;
while (count < n) {
int cur = vec[count].second;
if (cur > left && cur < right) {
count++;
continue;
}

if (cur < left) {
for (int i=cur+1; i<left; i++) {
total += (vec[count].first - A[i]);
}
left = cur;
}

if (cur > right) {
for (int i=right+1; i<cur; i++) {
total += (vec[count].first - A[i]);
}
right = cur;
}

count++;
}

return total;
}
};