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.前綴法。第一個優化是把連續和轉變為sum[j] - sum[i]的值。第二個優化是引入前綴，一直記錄前i個數裏最小的sum[i]，那你每次算i~j和的時候肯定是要減去這個最小前綴才能讓差值最大化的。

2.貪婪法。每次如果發現當前的sum變成負數了就立即舍去，下一個數加進來的時候肯定選擇不要前面這串累贅的。

1.前綴法實現

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.貪心法實現

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