[LintCode] 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.

 

Example

Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

Challenge

O(n) time and O(1) memory

O(n) time and O(n) memory is also acceptable.

class Solution {
public:
    int trap(vector<int>& heights) {
        if(heights.size() < ) return ;
        int waterNum = , left = , right = heights.size() - ,
            leftH = heights[left], rightH = heights[right];

        while(left < right){
            if(leftH < rightH){
                int t = left + ;
                if(t < right){
                    if(heights[t] < leftH)
                        waterNum += leftH - heights[t];
                    else leftH = heights[t];
                }
                left = t;
            }else{
                int t = right - ;
                if(t > left){
                    if(heights[t] < rightH)
                        waterNum += rightH - heights[t];
                    else rightH = heights[t];
                }
                right = t;
            }
        }
        return waterNum;
    }
};