leetcode042：Trapping Rain Water
來(lái)源：程序員人生 發(fā)布時(shí)間：2015-08-13 08:11:56 閱讀次數(shù)：3363次
問(wèn)題描寫(xiě)
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!
問(wèn)題分析
這道題目解題思路可以參考leetcode011：Container With Most Water 的分析，leetcode011是求圍住的最大面積，這里有所區(qū)分，統(tǒng)計(jì)“裝水”面積，還是從兩邊向中間掃描，時(shí)間復(fù)雜度O(n)。 這里設(shè)置1個(gè)水平面h，表示當(dāng)前圍住的高度，如果后續(xù)掃面到的比h低，說(shuō)明可以裝水，統(tǒng)計(jì)；如果比h高，更新h便可。
代碼
//運(yùn)行時(shí)間：13ms
class Solution {
public:
int trap(vector<int>& height) {
int i = 0, j = height.size()⑴;
int ans = 0;
int h = 0;
while (j-i >= 0){
if (height[i] > height[j]){
if (height[j] <= h){
ans += (h - height[j]);
}
else{
h = height[j];
}
j--;
}
else if (height[i] < height[j]){
if (height[i] <= h){
ans += (h - height[i]);
i++;
}
else{
h = height[i];
}
}
else{
if (height[i] <= h){
if (i != j)
ans += 2 * (h - height[i]);
else
ans += (h - height[i]);
}
else{
h = height[i];
}
i++; j--;
}
}
return ans;
}
};
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