a5abef3b2d
* Create 11. Container With Most Water * Rename 11. Container With Most Water to containerwithmostwater.java * Update README.md
28 lines
1.1 KiB
Java
28 lines
1.1 KiB
Java
// Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai).
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// n vertical lines are drawn such that the two endpoints of the line i is at (i, ai) and (i, 0).
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// Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.
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// Notice that you may not slant the container.
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// Solution: Start with walls at either end. If the current area is greater than the max area, update it.
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// At each step, move the shorter wall one step towards the longer wall and recalculate. O(n) complexity.
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class Solution {
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public int maxArea(int[] height) {
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List<Integer> list = new ArrayList<Integer>();
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int start = 0, end = height.length - 1, multiplier = height.length - 1;
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int maxarea = Integer.MIN_VALUE;
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while (multiplier != 0) {
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maxarea = Integer.max(maxarea, Integer.min(height[start], height[end]) * multiplier--);
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if (height[start] > height[end]) {
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end--;
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} else {
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start++;
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}
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}
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return maxarea;
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}
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}
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