[LeetCode]Trapping Rain Water，leetcodetrapping

[LeetCode]Trapping Rain Water，leetcodetrapping

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.

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!
題意：求由上面的數組組成的bar所組成的槽所能夠容納的水量，
思路1：首先找到最大的bar然後分別從左向右計算盛水量，計算時先選定一個非0的bar然後如果後面的bar不比當前大，則依次入棧，直到找到大於當前的，然後對站內的bar依次計算盛水量，最後從右向左重複上面的步驟。
代碼1：

    public int trap1(List<Integer> height) {//兩邊向最長的bar掃面遇見比當前小的就入棧，直到遇到大的就彈出計算盛水量        int sum = 0;        int highIndex = 0;        int max = Integer.MIN_VALUE;        for(int i = 0; i < height.size(); ++ i){            if(height.get(i) > max){                max = height.get(i);                highIndex = i;            }        }        int left;        Stack<Integer> s = new Stack<Integer>();        for(int i = 0; i < highIndex;){            while (i < highIndex && height.get(i) == 0){                i ++;            }            left = height.get(i);            while (i< highIndex) {                while (i < highIndex && height.get(i) <= left) {                    s.push(height.get(i));                    i++;                }                while (!s.isEmpty()) {                    int temp = s.peek();                    s.pop();                    sum += (left - temp);                }                left = height.get(i);                i ++;            }        }        int right = 0;        for(int i = height.size() - 1; i > highIndex;){            while (i > highIndex && height.get(i) == 0){                i --;            }            right = height.get(i);            while (i > highIndex){                while (i > highIndex && height.get(i) <= right){                    s.push(height.get(i));                    i --;                }                while (!s.isEmpty()){                    sum += right - s.peek();                    s.pop();                }                right = height.get(i);                i --;            }        }        return sum;    }

思路2：和上面的計算思路類似，這次不必要首先計算最大的bar,而是每次比較左右兩邊選定的bar的大小，小的那個先計算，依次迴圈。空間O(1),時間O(N)
代碼：

    public int trap(List<Integer> height){        int sum = 0;        int left = 0, right = height.size() - 1;        while (left < right && height.get(left) == 0){            left ++;        }        while (right > left && height.get(right) == 0){            right -- ;        }        int leftHeight,rightHeight;        while (left < right) {            leftHeight = height.get(left);            rightHeight = height.get(right);            if (leftHeight <= rightHeight ){//當左邊bar高度小於右邊的高度的時候，左邊先掃描，直到左邊遇到比leftHeight 高為止                while (left < right && height.get(left) <= leftHeight){                    sum += leftHeight - height.get(left);                    left ++;                }            }else {                while (left < right && height.get(right) <= rightHeight ) {                    sum += rightHeight  - height.get(right);                    right--;                }            }        }        return sum;    }