最大子数组（最小子数组）

Given an array of integers, find a contiguous subarray which has the largest sum.
Return the sum of the subarray.

给出数组[−2,2,−3,4,−1,2,1,−5,3]，符合要求的子数组为[4,−1,2,1]，其最大和为6

1、暴力 O(n2)

public int maxSubArray(int[] nums) {
    if (nums == null || nums.length == 0) {
        return 0;
    }
    int maxSum = Integer.MIN_VALUE;
    for(int i=0; i<nums.length; i++){
        int sum = 0; //必须定义在第一个for内部
        for(int j=i; j<nums.length; j++){
            sum += nums[j];
            maxSum = Math.max(maxSum, sum);
        }
    }
    return maxSum;
}

2、贪婪算法

public int maxSubArray(int[] A) {
    if(A==null || A.length==0){
        return 0;
    }
    int maxSum = Integer.MIN_VALUE;
    int sum = 0;
    for(int i=0; i<A.length; i++){
        sum += A[i];
        maxSum = Math.max(maxSum, sum);
        sum = Math.max(sum, 0);//最关键 --sum 为负数时，令sum为0
    }
    return maxSum;
}

3、同方法2一样

public int maxSubArray(int[] nums) {
    int maxsum = Integer.MIN_VALUE;
    int sum = 0;
    for (int i = 0; i < nums.length; i++) {
        if (sum < 0) {
            sum = 0;
        }
        sum += nums[i];
        maxsum = Math.max(maxsum, sum);
    }
    return maxsum;
}

4、动态规划

public int maxSubArray(int[] nums) {
    int max_ending_here = nums[0]; //包括当前位置时候的最大值
    int max_so_far = nums[0]; // 最大值
    for (int i = 1; i < nums.length; i++) {
        max_ending_here = Math.max(nums[i], nums[i] + max_ending_here);
        max_so_far = Math.max(max_so_far, max_ending_here);
    }
    return max_so_far;
}

最小子数组

public int minSubArray(ArrayList<Integer> nums) {
    int min_ending_here = nums.get(0);
    int min_so_far = nums.get(0);
    for (int i=1 ;i< nums.size(); i++){
        min_ending_here = Math.min(nums.get(i),nums.get(i) + min_ending_here);
        min_so_far = Math.min( min_so_far, min_ending_here );
    }
    return min_so_far;
}