#363 Trapping Rain Water

题目描述：

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.

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Example

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

题目思路：

这题的精髓在于：两根柱子能围多少水，是由最短的柱子决定的。根据这个，就不难想到用two pointers的方法，一根l指针在最左，一根r指针在最右，那么它们实际的valid height = min(heights[l], heights[r])。这个height是一个一直被维护的值：当heights[l] < heights[r]时，如果heights[l] < height，那就意味着l的地方可以蓄水，蓄水量为height - heights[l]，此时height就不用更新，挪动l指针l++就可以进入下一轮; 如果heights[l] > height，那么此时无法蓄水，但是可以得到一个新的height = min(heights[l], heights[r]).当heights[l] >= heights[r]时，情况类似。

Mycode （AC = 33ms）：

class Solution {public:    /**     * @param heights: a vector of integers     * @return: a integer     */    int trapRainWater(vector<int> &heights) {        // write your code here        if (heights.size() <= 2) {            return 0;        }                // l = 0, r = -1        int l = 0, r = heights.size() - 1;        int height = min(heights[l], heights[r]);        int water = 0, delta = 0;;                while (l < r) { // 1 vs. 3, height = 2, water = 2 + 2 + 1            if (heights[l] < heights[r]) {                delta = heights[l] < height?                         height - heights[l] : 0; // 0                water += delta;                height = min(heights[r], max(heights[l], height)); // 2                                l++;            }            else {                delta = heights[r] < height?                         height - heights[r] : 0; // 0                water += delta;                height = min(heights[l], max(heights[r], height)); // height = 1                r--;            }        }                return water;    }};