标签：div 最大 一个 min contain logs code val 贪心
Find the contiguous subarray within an array (containing at least one number) which has the largest sum.
For example, given the array [-2,1,-3,4,-1,2,1,-5,4],
the contiguous subarray [4,-1,2,1] has the largest sum = 6.
1.前缀法。第一个优化是把连续和转变为sum[j] - sum[i]的值。第二个优化是引入前缀，一直记录前i个数里最小的sum[i]，那你每次算i~j和的时候肯定是要减去这个最小前缀才能让差值最大化的。
2.贪婪法。每次如果发现当前的sum变成负数了就立即舍去，下一个数加进来的时候肯定选择不要前面这串累赘的。
1.前缀法实现
class Solution { public int maxSubArray(int[] nums) { if (nums == null || nums.length == 0) { return 0; } int[] sums = new int[nums.length + 1]; sums[0] = 0; for (int i = 0; i < nums.length; i++) { sums[i + 1] = sums[i] + nums[i]; } int minPreSum = sums[0]; int maxSum = Integer.MIN_VALUE; for (int i = 1; i < sums.length; i++) { int crtSum = sums[i] - minPreSum; maxSum = Math.max(maxSum, crtSum); minPreSum = Math.min(minPreSum, sums[i]); } return maxSum; } }
2.贪心法实现
class Solution { public int maxSubArray(int[] nums) { if (nums == null || nums.length == 0) { return 0; } int maxSum = Integer.MIN_VALUE; int localSum = 0; for (int i = 0; i < nums.length; i++) { localSum += nums[i]; maxSum = Math.max(maxSum, localSum); localSum = Math.max(localSum, 0); } return maxSum; } }
leetcode53- Maximum Subarray- easy
标签：div 最大 一个 min contain logs code val 贪心
原文地址：http://www.cnblogs.com/jasminemzy/p/7818815.html
