[leetcode-42]Trapping Rain Water(java)

问题描述：
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.

分析：我的解题思路是和图中的所示面积是相吻合的。这里面维护一个堆栈，如果当前元素大于栈顶元素，出栈，并在两个元素中维护一个面积数组。然后再进行完之后，处理一下余下的堆栈数据。然后再计算面积之和。

代码如下：344ms

public class Solution {
        public int trap(int[] height) {
        Stack heightStack = new Stack<>();
        List indexList = new LinkedList<>();
        int length = height.length;
        int[] area = new int[length];

        for(int i = 0;iif(heightStack.isEmpty()){
                heightStack.push(height[i]);
                indexList.add(i);
                continue;
            }
            int val = height[i];

            while(!heightStack.isEmpty()){
                int prev = heightStack.peek();
                if(valbreak;

                int prevIndex = indexList.remove(indexList.size() - 1);
                for(int j = prevIndex+1;jint head;
        int headIndex;

        while(!heightStack.isEmpty()){
            head = heightStack.pop();
            headIndex = indexList.remove(indexList.size()-1);

            if(heightStack.isEmpty())
                break;

            int peekIndex = indexList.get(indexList.size()-1);
            for(int i = headIndex-1;i>peekIndex;i--){
                area[i] = head-height[i];
            }
        }
        int count = 0;
        for(int i = 0;ireturn count;
    }
}