Leetcode Container with the most water (C,c++,java,python)

Problem:

GivenNnon-negative integersa1 ,a2 , ...,an , where each represents a point at coordinate (I,ai ).NVertical lines is drawn such that the both endpoints of lineIis at (I,ai ) and (I, 0). Find lines, which together with X-axis forms a container, such that the container contains the most water.

Note:you may not slant the container.

Solution:

The following example: [4,6,2,6,7,11,2] to explain.

1. First, suppose we find the longitudinal line that can take the maximum volume as I, J (assuming I<j), then the maximum volume C = min (ai, AJ) * (j-i);

2. Let's look at a property like this:

①: No line at the right end of J will be higher than it! Hypothesis exists K | (j<k && ak > AJ), then by ak> aj, so min (Ai,aj, AK) =min (AI,AJ), so the volume of the container consisting of I, k c ' = min (ai,aj) * (k-i) > C, and C is the most value contradiction, so the proof of J will not have a higher line than it;

②: Similarly, there will be no higher line on the left side of I;

What does that mean? If we currently get the candidate: set to X, y two lines (x< y), then be able to get a larger volume than its new two edges necessarily within the [x, Y] interval and ax ' > =ax, ay ' >= ay;

3. So we move from the two to the middle, while updating the candidate values, in the contraction interval priority from the X, y of the smaller edge of the contraction;

Title: On the X-axis given a number of vertical lines, the vertical line has a length, to find two vertical and x-axis a container can accommodate the most water area problem-solving ideas: such as solution, Time complexity O (n) Java source code (spents 373ms):

public class Solution {public    int maxarea (int[] height) {        int max=-1,l=0,r=height.length-1;        while (l<r) {            int area= (Height[l]

C Language Source code (spents 12ms):
int Maxarea (int* height, int heightsize) {    int max=-1,area,l=0,r=heightsize-1,k;    while (l<r) {        area= (Height[l]
C + + source code (spents 30ms):
Class Solution {public:    int Maxarea (vector<int>& height) {        int max=-1,l=0,r=height.size () -1,area,k ;        while (l<r) {            area= (Height[l]
Python source code (spents 152ms):
Class solution:    # @param {integer[]} height    # @return {integer}    def maxarea (self, height):        max=-1;l=0; R=len (height)-1 while        l<r:            area= (Height[l] if height[l]



Leetcode Container with the most water (C,c++,java,python)