[leetcode 42] Trapping Rain Water

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：首先找到最高的柱子i, 用最高的柱子将其划分为左右两个部分。

class Solution {
public:
int trap(int A[], int n) {
int res = 0;
int highest = -1;
int index = -1;
for (int i = 0; i < n; i++) {
if (highest < A[i]) {
highest = A[i];
index = i;
}
}
for (int i = 0, peek = 0; i < index; i++) {
//当前位置储水
if (peek > A[i]) {
res += peek - A[i];
}
//i+1位置左边最高点，只有左右都高，此处才能蓄水
peek = max(peek, A[i]);
}
for (int i = n - 1, peek = 0; i > index; i--) {
if (peek > A[i]) {
res += peek - A[i];
}
peek = max(peek, A[i]);
}
return res;
}
};