【leetcode每日刷题】【DP】42. Trapping Rain Water

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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.

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!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

【DP】

图示：1、先获得从开始到当前index的bar最大高度；2、获得从末尾节点开始到当前index的最大bar的高度。3、将 min(max_left,max_right)-height，进行累加。

class num42 {
    public int trap(int[] height) {
        if(height.length == 0) return 0;
        int ans = 0;
        int size = height.length;
        int[] max_left = new int[size];
        int[] max_right = new int[size];
        max_left[0] = height[0];
        for(int i=1; i<size; i++){
            max_left[i] = Math.max(height[i], max_left[i-1]);
        }
        max_right[size-1] = height[size-1];
        for(int i=size-2; i>=0; i--){
            max_right[i] = Math.max(height[i], max_right[i+1]);
        }
        for(int i=1; i<size; i++){
            ans += Math.min(max_left[i], max_right[i])-height[i];
        }
        return ans;
    }
    public static void main(String[] args) {
        int[] height = {0,1,0,2,1,0,1,3,2,1,2,1};
        num42 solution = new num42();
        System.out.println(solution.trap(height));
    }
}