问题：

We are given an elevation map, heights[i] representing the height of the terrain at that index. The width at each index is 1. After Vunits of water fall at index K, how much water is at each index?

Water first drops at index K and rests on top of the highest terrain or water at that index. Then, it flows according to the following rules:

• If the droplet would eventually fall by moving left, then move left.
• Otherwise, if the droplet would eventually fall by moving right, then move right.
• Otherwise, rise at it's current position.
• Here, "eventually fall" means that the droplet will eventually be at a lower level if it moves in that direction. Also, "level" means the height of the terrain plus any water in that column.

  We can assume there's infinitely high terrain on the two sides out of bounds of the array. Also, there could not be partial water being spread out evenly on more than 1 grid block - each unit of water has to be in exactly one block.

  Example 1:

  Input: heights = [2,1,1,2,1,2,2], V = 4, K = 3
  Output: [2,2,2,3,2,2,2]
  Explanation:
  #       #
  #       #
  ##  # ###
  #########0123456    <- index
  The first drop of water lands at index K = 3:
  #       #
  #   w   #
  ##  # ###
  #########0123456
  When moving left or right, the water can only move to the same level or a lower level.
  (By level, we mean the total height of the terrain plus any water in that column.)
  Since moving left will eventually make it fall, it moves left.
  (A droplet "made to fall" means go to a lower height than it was at previously.)
  #       #
  #       #
  ## w# ###
  #########0123456
  Since moving left will not make it fall, it stays in place.  The next droplet falls:
  #       #
  #   w   #
  ## w# ###
  #########0123456
  Since the new droplet moving left will eventually make it fall, it moves left.
  Notice that the droplet still preferred to move left,
  even though it could move right (and moving right makes it fall quicker.)
  #       #
  #  w    #
  ## w# ###
  #########0123456
  #       #
  #       #
  ##ww# ###
  #########0123456
  After those steps, the third droplet falls.
  Since moving left would not eventually make it fall, it tries to move right.
  Since moving right would eventually make it fall, it moves right.
  #       #
  #   w   #
  ##ww# ###
  #########0123456
  #       #
  #       #
  ##ww#w###
  #########0123456
  Finally, the fourth droplet falls.
  Since moving left would not eventually make it fall, it tries to move right.
  Since moving right would not eventually make it fall, it stays in place:
  #       #
  #   w   #
  ##ww#w###
  #########0123456
  The final answer is [2,2,2,3,2,2,2]:#    ####### ####### 0123456

  Example 2:

  Input: heights = [1,2,3,4], V = 2, K = 2
  Output: [2,3,3,4]
  Explanation:
  The last droplet settles at index 1, since moving further left would not cause it to eventually fall to a lower height.

  Example 3:

  Input: heights = [3,1,3], V = 5, K = 1
  Output: [4,4,4]

  Note:

  1. heights will have length in [1, 100] and contain integers in [0, 99].
  2. V will be in range [0, 2000].
  3. K will be in range [0, heights.length - 1].

解决：

【题意】

给定一个组高度值，代表一个水槽的底部高度分布情况。在K点处倒入V体积的水，求倒水之后的高度分布。

倒入的每一体积的水按照如下规则进行流动：

1. 如果在K点左侧存在地势较低且可达的位置，则水优先向左流动。
2. 否则，在K点右侧存在地势较低且可达的位置，则水向右侧流动。
3. 否则，水停留在K处。

①  直接模拟上面的流程。

class Solution { //7ms
    public int[] pourWater(int[] heights, int V, int K) {
        for (int i = 0;i < V;i ++){
            int leftMinIndex = findLeftMinIndex(heights,K);
            if (leftMinIndex < K){
                heights[leftMinIndex] ++;
            }else {
                int rightMinIndex = findRightMinIndex(heights,K);
                if (rightMinIndex > K){
                    heights[rightMinIndex] ++;
                }else {
                    heights[K] ++;
                }
            }
        }
        return heights;
    }
    public int findLeftMinIndex(int[] heights,int K){
        int minIndex = K;
        int minHeight = heights[K];
        for (int i = K - 1;i >= 0;i --){
            if (heights[i] < minHeight){
                minIndex = i;
                minHeight = heights[i];
            }else if (heights[i] > minHeight){
                break;
            }
        }
        return minIndex;
    }
    public int findRightMinIndex(int[] heights,int K){
        int minIndex = K;
        int minHeight = heights[K];
        for (int i = K + 1;i < heights.length;i ++){
            if (heights[i] < minHeight){
                minIndex = i;
                minHeight = heights[i];
            }else if (heights[i] > minHeight){
                break;
            }
        }
        return minIndex;
    }
}