53.Maximum Subarray

    /** 53.Maximum Subarray * 2016-5-7 by Mingyang * If we traverse this array from the beginning.     For one of the elements in the array, it has only two choices: 1. * Either join the previous array Plus and in (with another group) * 2. Or you can set up an array of your own (one by one) * So how to choose this element depends on what group he will contribute to.     If you have a group with others * can make the total increase and get bigger, or with others a group of good, if you start a group, the value of their own than the previous sums value is still larger, then you open a single set of good. * So use a DP array to record the maximum value for each round of sum, Dp[i] Indicates whether the current element is preceded by an array plus a set or a single set of itself, and then maintains a global maximum of the ascended answer * then this topic starts to think can be used Maxsubarray (int a[], int i , Int j), which means the Maxsubarray for a[i:j]. * But it's hard to find this kind of relationship by writing a sub-function like this.     So what do we do next?     * Maxsubarray (int a[], int i), which means the Maxsubarray for A[0:I] which must have a[i] as the end element.     * Then the relationship is: * Dp[i] = Math.max (A[i], dp[i-1] + a[i]);     * That is to say, for A[i] in the end, we just need to see this add in big or alone big.     * Because if you add in is bigger than the individual, then the back is an increment.     * Never write: dp[i] = Math.max (Dp[i-1], dp[i-1] + a[i]); */     Public Static intMaxsubarray (int[] A) {int[] DP =New int[A.length]; intmax = A[0]; dp[0] = a[0];  for(inti = 1; i < a.length; i++) {Dp[i]= Math.max (A[i], dp[i-1] + a[i]);//here only compare their own open another, and the original plus open the differenceMax =Math.max (Max, dp[i]); }        returnMax; }    /** Below is my own solution, began to think a bit complex, and then a careful enumeration is quite simple, here with a DP array * Dp[i] to represent the maximum of continuous array including I, rather than to I optimal results * [ -1,-2] so dp[1]=-3, because To include the 2. * Then if each number itself is very large, than the previous cumulative DP is larger, so long playing independently * otherwise need to add up, each time the DP is updated with the global variable Max compared*/     Public intMaxSubArray1 (int[] nums) {        intlen=nums.length; intMax=nums[0]; int[] dp=New int[Len]; dp[0]=nums[0];  for(inti=1;i<len;i++){            if(nums[i]>nums[i]+dp[i-1]) {Dp[i]=Nums[i]; Max=Math.max (Max,dp[i]); }Else{Dp[i]=dp[i-1]+Nums[i]; Max=Math.max (Max,dp[i]); }        }        returnMax; }

53.Maximum Subarray