42. Trapping Rain Water

题目：
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.

解答：

public class Solution {
    public int trap(int[] height) {
        //左边比右边小或者大都可以盛水，所以我们不能直接确定右边是否会有一个柱子比较大，能盛所有现在积攒的水。
        //那么我们就找到中间最大的那个柱子，把它分成左右两边，那么不管从左边还是右边都能保证最后可以有最高的柱子在，之前盛的水都是有效的
        if (height.length <= 2) return 0;
        int maxHeight = 0, maxIndex = 0;
        int result = 0;
        //find the max height and its index
        for (int i = 0; i < height.length; i++) {
            if (height[i] > maxHeight) {
                maxHeight = height[i];
                maxIndex = i;
            }
        }
        //left part
        int maxLeft = height[0];
        for (int i = 1; i < maxIndex; i++) {
            if (height[i] > maxLeft) {
                maxLeft = height[i];
            } else {
                result += maxLeft - height[i];
            }
        }
        //right part
        int maxRight = height[height.length - 1];
        for (int i = height.length - 2; i > maxIndex; i--) {
            if (height[i] > maxRight) {
                maxRight = height[i];
            } else {
                result += maxRight - height[i];
            }
        }
        return result;
    }
}