leetcode 42. Trapping Rain Water
阿新 • • 發佈:2017-10-03
pre lan pub after lin ret src esc problem
link
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
.
題意:看文字比較奇怪,看圖片很清晰。給出每塊的高度,求出往上面倒水那坨東西裏可以存放多少水。
思路:
每個bar 能存多少水取決於它左邊的最高的bar的高度和右邊最高的bar的高度的較小的那個, 與其自身高度的差。
因此方法很簡單,從左到右掃一次可以算出每個bar左邊的最高高度。從右到左掃一次可以算出每個bar右邊的最高的高度。
code:
class Solution { public: int trap(vector<int>& height) { if(height.size() == 0) return 0; vector<int>leftMax(height.size()); leftMax[0] = height[0]; for(int i = 1 ; i < height.size(); i++){ leftMax[i]= max(leftMax[i-1], height[i]); } int rightMax = INT_MIN; int ans = 0; for(int i = height.size() - 1; i >= 0; i--){ rightMax = max(rightMax, height[i]); ans += (min(rightMax, leftMax[i]) - height[i]); } return ans; } };
leetcode 42. Trapping Rain Water