【leetcode】42. Trapping Rain Water

题目如下：

解题思路：我的方法是找出所有的顶点，即满足height[i] > height[i+1] && height[i] > height[i-1]条件的index,如果height[i] == height[i+1],则需要往后顺延，找到i后面第一个不与自己相等的点，把所有的顶点存入列表peaks，接下来遍历peaks，对于任意一个peaks[i]，找出第一个j满足j > i && height[peaks[i]] > height[peaks[j]]，这一个区间就是一个盛水区间，如果找不到相应的j，那么就找出height[peaks[i] -> end] 内height的最大值。找出所有盛水区间后，计算总量即可。

代码如下：

class Solution(object):
    def getNextPeak(self,inx,peaks,height):
        maxH = 0
        maxInx = 0
        for i in range(inx+1,len(peaks)):
            if height[peaks[inx]] <= height[peaks[i]]:
                return i
            if maxH < height[peaks[i]]:
                maxH = height[peaks[i]]
                maxInx = i
        return maxInx
    def getNextNoDupHeightInx(self,inx,height):
        for i in range(inx,len(height)):
            if height[inx] != height[i]:
                return i
        return -1
    
    def getPeaks(self, peaks, height):
        i = 0
        while i < len(height):
            nextInx = self.getNextNoDupHeightInx(i, height)
            if nextInx == -1:
                if height[i] > height[i - 1]:
                    peaks.append(i)
                break
            if i == 0 and height[i] > height[nextInx]:
                peaks.append(i)
            elif height[i] > height[i - 1] and height[i] > height[nextInx]:
                peaks.append(i)
            i = nextInx

    def trap(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        if len(height) < 3:
            return  0
        peaks = []
        self.getPeaks(peaks,height)
        print peaks
        res = 0
        i = 0
        while i < len(peaks)-1:
            nextInx = self.getNextPeak(i,peaks,height)
            high = min(height[peaks[i]],height[peaks[nextInx]])
            for j in xrange(peaks[i],peaks[nextInx]):
                water = max(0,high - height[j])
                res += water
            i = nextInx
        return res