Algorithms—42.Trapping Rain Water

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public class Solution {        public int trap(int[] height) {    if (height.length<=1) {return 0;}    int a=0;    int h=height[a];    /**     * 求出最高峰     */        for (int i = 1; i < height.length; i++) {        if (height[i]>h) {a=i;h=height[i];}}        Map<String, Integer> leftMap=new HashMap<String, Integer>();        Map<String, Integer> rightMap=new HashMap<String, Integer>();        leftMap.put("result", 0);        leftMap.put("a",a);        rightMap.put("result", 0);        rightMap.put("a",a);        Map<String, Integer> leftrRsult=new Solution().left(leftMap, height);        Map<String, Integer> rightResult=new Solution().right(leftMap, height);    return leftrRsult.get("result")+rightResult.get("result");    }        /**     * 求左侧     * @param height     * @param a     * @return     */    public Map<String, Integer> left(Map<String, Integer> m,int[] height){    Map<String, Integer> map=new HashMap<String, Integer>();    if (m==null||m.get("a")==null) {return m;}    if (m.get("a")<=1) {return m;}    int h=height[0];    int result=m.get("result");    int sum=0;    int a=0;    for (int i = 1; i <m.get("a"); i++) {if (height[i]<h) {sum+=(h-height[i]);}else {sum=0;h=height[i];a=i;}}    result+=sum;    map.put("result", result);    map.put("a", a);                return left(map, height);    }    /**     * 求右侧     * @param height     * @param a     * @return     */    public Map<String, Integer> right(Map<String, Integer> m,int[] height){    Map<String, Integer> map=new HashMap<String, Integer>();    if (m==null||m.get("a")==null) {return m;}    if (m.get("a")>=height.length-2) {return m;}    int h=height[height.length-1];    int result=m.get("result");    int sum=0;    int a=0;    for (int i = height.length-1; i>m.get("a"); i--) {if (height[i]<h) {sum+=(h-height[i]);}else {sum=0;h=height[i];a=i;}}    result+=sum;    map.put("result", result);    map.put("a", a);                return right(map, height);    }}

思路：先找出最高峰，然后分为左右两个部分，以左边为例说明，从开始端作为参照开始往右遍历，如果遇到小于参照值的，那么加上参照值与当前值的差值，否则更换参照值；求出靠近最高峰最大值之间的蓄水量，然后递归进行再次求解。右边同理。

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